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incompatible pointer type

P: n/a
code segment:

long int * size;
char entry[4][16];
.............
size=&entry[row][col];

*************************************
Gcc reported 'assignment of incompatible pointer type'.

MY QUESTION IS:
I understand that pointer is just a memory unit that contains the
address(or starting address)of other data object(i.e. int, float,
array, struct, etc.). So it is just an address. How does C implement
the pointer underlying? Why there can be different pointer types. If
so, what is the difference between an integer pointer type and a float
pointer type, for example?

Thank you!

Ji

Mar 7 '07 #1
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13 Replies


P: n/a

"william" <wi**********@gmail.comwrote in message
news:11**********************@c51g2000cwc.googlegr oups.com...
code segment:

long int * size;
char entry[4][16];
............
size=&entry[row][col];

*************************************
Gcc reported 'assignment of incompatible pointer type'.
The report is correct.
You should change 'size' to type 'char *'
>
MY QUESTION IS:
I understand that pointer is just a memory unit that contains the
address(or starting address)of other data object(i.e. int, float,
array, struct, etc.). So it is just an address. How does C implement
the pointer underlying?
That's specific to the implementation.
>Why there can be different pointer types.
Because the language defines them as different.
If
so, what is the difference between an integer pointer type and a float
pointer type, for example?
The difference is that they're two different types.

-Mike
Mar 7 '07 #2

P: n/a
"william" <wi**********@gmail.comwrote in message
news:11**********************@c51g2000cwc.googlegr oups.com...
code segment:

long int * size;
char entry[4][16];
............
size=&entry[row][col];

*************************************
Gcc reported 'assignment of incompatible pointer type'.

MY QUESTION IS:
I understand that pointer is just a memory unit that contains the
address(or starting address)of other data object(i.e. int, float,
array, struct, etc.). So it is just an address. How does C implement
the pointer underlying? Why there can be different pointer types. If
so, what is the difference between an integer pointer type and a float
pointer type, for example?

Thank you!

Ji
And int pointer type and a float pointer type point to different types of
data. int* points to data that contains an integer. float* points to data
that contains a float. There may be some circumstances where you may wish
to convert between pointer types, but you better know what you're doing.

In your example, you are making size a pointer to an int, but then trying to
get it to point to a character. The compiler rightfully says, wait a second
buddy, those are two different types. If you really wish to do this, the
you could use reinterpret_cast to change the type of the pointer. It would
compile but depending on hwo you are use it could cause all kinds of
problems at run time. I.E.

size = reinterpret_cast<int*>( &entry[row][col] );

Again, this is very dangerous. Especially since a char is only 1 character
and an interger is more (4 on my system, 8 on some, who knows how many on
others).
Mar 7 '07 #3

P: n/a
S S
On Mar 7, 8:13 pm, "william" <william.m...@gmail.comwrote:
code segment:

long int * size;
char entry[4][16];
............
size=&entry[row][col];

*************************************
Gcc reported 'assignment of incompatible pointer type'.

MY QUESTION IS:
I understand that pointer is just a memory unit that contains the
address(or starting address)of other data object(i.e. int, float,
array, struct, etc.). So it is just an address. How does C implement
the pointer underlying? Why there can be different pointer types. If
so, what is the difference between an integer pointer type and a float
pointer type, for example?

Thank you!

Ji
Say, how would you increment the pointer then? You need to do several
operations on pointer and if compiler does not know what type it is,
you cannt do even p++. As simple as that.

Mar 7 '07 #4

P: n/a
On Mar 7, 11:32 am, "S S" <sarvesh.si...@gmail.comwrote:
On Mar 7, 8:13 pm, "william" <william.m...@gmail.comwrote:
code segment:
long int * size;
char entry[4][16];
............
size=&entry[row][col];
*************************************
Gcc reported 'assignment of incompatible pointer type'.
MY QUESTION IS:
I understand that pointer is just a memory unit that contains the
address(or starting address)of other data object(i.e. int, float,
array, struct, etc.). So it is just an address. How does C implement
the pointer underlying? Why there can be different pointer types. If
so, what is the difference between an integer pointer type and a float
pointer type, for example?
Thank you!
Ji

Say, how would you increment the pointer then? You need to do several
operations on pointer and if compiler does not know what type it is,
you cannt do even p++. As simple as that.
I understand it. It makes a lot of sense. Pointer++ :-)

Mar 7 '07 #5

P: n/a
On Mar 7, 3:12 pm, "william" <william.m...@gmail.comwrote:
On Mar 7, 11:32 am, "S S" <sarvesh.si...@gmail.comwrote:
On Mar 7, 8:13 pm, "william" <william.m...@gmail.comwrote:
code segment:
long int * size;
char entry[4][16];
............
size=&entry[row][col];
*************************************
Gcc reported 'assignment of incompatible pointer type'.
MY QUESTION IS:
I understand that pointer is just a memory unit that contains the
address(or starting address)of other data object(i.e. int, float,
array, struct, etc.). So it is just an address. How does C implement
the pointer underlying? Why there can be different pointer types. If
so, what is the difference between an integer pointer type and a float
pointer type, for example?
Thank you!
Ji
Say, how would you increment the pointer then? You need to do several
operations on pointer and if compiler does not know what type it is,
you cannt do even p++. As simple as that.

I understand it. It makes a lot of sense. Pointer++ :-)
**********************************
My original intention was to use any bytes to comprise an int. So I
used two ways to do that:

1.the function atoi():

However, I still get some problems here:
char c[]="ABCD";
int i;

i=atoi(&c[0]);// I expected to get 65: A's ascii value

the result of the line of code above is ZERO, which confuse me a lot.

2.conversion:

i=(int)c[0];

this case it worked. However, if c[0]is arbitery. 'i' always get
negative value, how come?

Thanks
Mar 7 '07 #6

P: n/a
On Mar 7, 2:19 pm, "william" <william.m...@gmail.comwrote:
My original intention was to use any bytes to comprise an int. So I
used two ways to do that:

1.the function atoi():

However, I still get some problems here:
char c[]="ABCD";
int i;

i=atoi(&c[0]);// I expected to get 65: A's ascii value

the result of the line of code above is ZERO, which confuse me a lot.
I think you are misinterpreting the meaning of atoi().

atoi() assumes that the string is an ASCII representation of an
integer.

char *c = "1234";
i = atoi(c); // i == 1234

When the string is not an ASCII representation of an integer (like
"ABCD"), it returns zero.

char *c = "ABCD";
i = atoi(c); // i == 0
>
2.conversion:

i=(int)c[0];

this case it worked. However, if c[0]is arbitery. 'i' always get
negative value, how come?
What do you mean by arbitrary?

Think of it this way -- the integral types in C/C++ are just that:
integers.
On most 32-bit architectures, a char is one byte, and an int is 4.
It's still a one-byte integer, though.

Mar 7 '07 #7

P: n/a
On Mar 7, 5:01 pm, jlongstr...@gmail.com wrote:
On Mar 7, 2:19 pm, "william" <william.m...@gmail.comwrote:
My original intention was to use any bytes to comprise an int. So I
used two ways to do that:
1.the function atoi():
However, I still get some problems here:
char c[]="ABCD";
int i;
i=atoi(&c[0]);// I expected to get 65: A's ascii value
the result of the line of code above is ZERO, which confuse me a lot.

I think you are misinterpreting the meaning of atoi().

atoi() assumes that the string is an ASCII representation of an
integer.

char *c = "1234";
i = atoi(c); // i == 1234

When the string is not an ASCII representation of an integer (like
"ABCD"), it returns zero.

char *c = "ABCD";
i = atoi(c); // i == 0
2.conversion:
i=(int)c[0];
this case it worked. However, if c[0]is arbitery. 'i' always get
negative value, how come?

What do you mean by arbitrary?
By arbitrary, I mean if c[0] contains any possible valued raning from
0x00~0xff. Then sometimes, (int)any_char returns negative value.

I got a solution now: if I specify "any_char" as an unsigned char, the
results of conversion is always postive.

Again, does any one has better solution to creat an int using 4
individual bytes(or a 4 byte array).

Thanks
Think of it this way -- the integral types in C/C++ are just that:
integers.
On most 32-bit architectures, a char is one byte, and an int is 4.
It's still a one-byte integer, though.

Mar 8 '07 #8

P: n/a
william wrote:
On Mar 7, 5:01 pm, jlongstr...@gmail.com wrote:
>>On Mar 7, 2:19 pm, "william" <william.m...@gmail.comwrote:

>>>My original intention was to use any bytes to comprise an int. So I
used two ways to do that:
>>>1.the function atoi():
>>>However, I still get some problems here:
char c[]="ABCD";
int i;
>>>i=atoi(&c[0]);// I expected to get 65: A's ascii value
>>>the result of the line of code above is ZERO, which confuse me a lot.

I think you are misinterpreting the meaning of atoi().

atoi() assumes that the string is an ASCII representation of an
integer.

char *c = "1234";
i = atoi(c); // i == 1234

When the string is not an ASCII representation of an integer (like
"ABCD"), it returns zero.

char *c = "ABCD";
i = atoi(c); // i == 0

>>>2.conversion:
>>>i=(int)c[0];
>>>this case it worked. However, if c[0]is arbitery. 'i' always get
negative value, how come?

What do you mean by arbitrary?

By arbitrary, I mean if c[0] contains any possible valued raning from
0x00~0xff. Then sometimes, (int)any_char returns negative value.

I got a solution now: if I specify "any_char" as an unsigned char, the
results of conversion is always postive.

Again, does any one has better solution to creat an int using 4
individual bytes(or a 4 byte array).

Thanks
>>Think of it this way -- the integral types in C/C++ are just that:
integers.
On most 32-bit architectures, a char is one byte, and an int is 4.
It's still a one-byte integer, though.


A union is one possiblilty

union S
{
char a[4];
int i;
};

S s;
s.a[0] = 'A';
s.a[1] = 'B';
s.a[2] = 'C';
s.a[3] = 'D';
cout << s.i;

Of course the are all sorts of platform dependencies here, this is not
portable code.

john
Mar 8 '07 #9

P: n/a
>
A union is one possiblilty

union S
{
char a[4];
int i;
};

S s;
s.a[0] = 'A';
s.a[1] = 'B';
s.a[2] = 'C';
s.a[3] = 'D';
cout << s.i;

Of course the are all sorts of platform dependencies here, this is not
portable code.

john
BTW i'm not trying to suggest the any sort of conversion from hex is
going on here (not sure if that is what you want or not).

john
Mar 8 '07 #10

P: n/a
(Received in e-mail, replying in group so people can correct any mistakes I
make)

On Mar 7, 10:56 am, "Jim Langston" <tazmas...@rocketmail.comwrote:
"william" <william.m...@gmail.comwrote in message

news:11**********************@c51g2000cwc.googlegr oups.com...
code segment:
long int * size;
char entry[4][16];
............
size=&entry[row][col];
*************************************
Gcc reported 'assignment of incompatible pointer type'.
MY QUESTION IS:
I understand that pointer is just a memory unit that contains the
address(or starting address)of other data object(i.e. int, float,
array, struct, etc.). So it is just an address. How does C implement
the pointer underlying? Why there can be different pointer types. If
so, what is the difference between an integer pointer type and a float
pointer type, for example?
Thank you!
Ji

And int pointer type and a float pointer type point to different types of
data. int* points to data that contains an integer. float* points to
data
that contains a float. There may be some circumstances where you may wish
to convert between pointer types, but you better know what you're doing.

In your example, you are making size a pointer to an int, but then trying
to
get it to point to a character. The compiler rightfully says, wait a
second
buddy, those are two different types. If you really wish to do this, the
you could use reinterpret_cast to change the type of the pointer. It
would
compile but depending on hwo you are use it could cause all kinds of
problems at run time. I.E.

size = reinterpret_cast<int*>( &entry[row][col] );

Again, this is very dangerous. Especially since a char is only 1
character
and an interger is more (4 on my system, 8 on some, who knows how many on
others).

- Thank you, Jim. What I really plan to do is to assign 4 arbitery bytes
- to an int. I was actually manipulating the bytes in MBR entry, I want
- to read the 4 bytes start from any address I specified(here I use the
- char[] to specify the address).
-
- So how could I get any 4 bytes in the MBR entry and convert it as an
- integer? Thank you again.
-
- Sincerely Ji

size = reinterpret_cast<int*>( somecharpointer );
*may* work for you, or it may not It depends on a lot on the architecture
you plan on running this on.

Some CPUs have problems reading integers that are not aligned on specific
byte boundaries, some do not. I believe (but could be mistaken) that AMD
and Intel are okay with this.

So, say, you had some char pointer pointing to arbitrary data that you can
read. If you are not worried about cross platform compatability, then I
would just point the int pointer to the start of where you think the array
is.

char* Data = SomeFunctionReturningChar*( someparm );
int* IntData = reinterpret_cast<int*>( Data );
At this point, if your architecture isn't too restrictive, you should be
able to read the contents of IntData as an integer, I.E.
std::cout << *IntData;
Should give you some int value, as long as the pointer points to memory you
have rights to read.
Notice, however, that incrementing your int pointer will increment it 4
bytes, not 1, because the compiler thinks it's int data you're pointing to.

++IntData;
will make IntData point to 4 bytes later, not one. Just as
IntData[1];
will look at the 5th through 8th bytes in the data.

Now, there are other ways to do it. There've been times I wanted to look at
data and didn't care to store it, so
std::cout << *reinterpret_cast<int*>( Data );
would give me an int value.
std::cout << *reinterpret_cast<int*>( Data + 1 );
would give me the 2nd through 5th bytes as an int, as would
std::cout << *reinterpret_cast<int*>( &Data[1] );

It really depends on what you are trying to accomplish how you would do it.
Looking and displaying the data the worst that usually can happen is you can
crash your program, reading data you don't own, etc.. Changing the data can
get you in trouble if you're not careful at what your'e pointing at, but
that's pretty much the same for all pointers.
Mar 8 '07 #11

P: n/a
John Harrison <jo*************@hotmail.comwrote:
A union is one possiblilty

union S
{
char a[4];
int i;
};

S s;
s.a[0] = 'A';
s.a[1] = 'B';
s.a[2] = 'C';
s.a[3] = 'D';
cout << s.i;
Technically it is undefined behavior to store something in a union and
then to try to read a different member of the union.
Of course the are all sorts of platform dependencies here, this is not
portable code.
Right, it *may* work on the OP's implementation.

--
Marcus Kwok
Replace 'invalid' with 'net' to reply
Mar 13 '07 #12

P: n/a
On Mar 13, 3:17 pm, ricec...@gehennom.invalid (Marcus Kwok) wrote:
John Harrison <john_androni...@hotmail.comwrote:
A union is one possiblilty
union S
{
char a[4];
int i;
};
S s;
s.a[0] = 'A';
s.a[1] = 'B';
s.a[2] = 'C';
s.a[3] = 'D';
cout << s.i;

Technically it is undefined behavior to store something in a union and
then to try to read a different member of the union.
Of course the are all sorts of platform dependencies here, this is not
portable code.

Right, it *may* work on the OP's implementation.

--
Marcus Kwok
Replace 'invalid' with 'net' to reply
Thank you for all the replies and the effort.

Mar 17 '07 #13

P: n/a
On Mar 8, 5:37 am, "Jim Langston" <tazmas...@rocketmail.comwrote:
(Received in e-mail, replying in group so people can correct any mistakes I
make)

On Mar 7, 10:56 am, "Jim Langston" <tazmas...@rocketmail.comwrote:
"william" <william.m...@gmail.comwrote in message
news:11**********************@c51g2000cwc.googlegr oups.com...
code segment:
long int * size;
char entry[4][16];
............
size=&entry[row][col];
*************************************
Gcc reported 'assignment of incompatible pointer type'.
MY QUESTION IS:
I understand that pointer is just a memory unit that contains the
address(or starting address)of other data object(i.e. int, float,
array, struct, etc.). So it is just an address. How does C implement
the pointer underlying? Why there can be different pointer types. If
so, what is the difference between an integer pointer type and a float
pointer type, for example?
Thank you!
Ji
And int pointer type and a float pointer type point to different types of
data. int* points to data that contains an integer. float* points to
data
that contains a float. There may be some circumstances where you may wish
to convert between pointer types, but you better know what you're doing.
In your example, you are making size a pointer to an int, but then trying
to
get it to point to a character. The compiler rightfully says, wait a
second
buddy, those are two different types. If you really wish to do this, the
you could use reinterpret_cast to change the type of the pointer. It
would
compile but depending on hwo you are use it could cause all kinds of
problems at run time. I.E.
size = reinterpret_cast<int*>( &entry[row][col] );
Again, this is very dangerous. Especially since a char is only 1
character
and an interger is more (4 on my system, 8 on some, who knows how many on
others).

- Thank you, Jim. What I really plan to do is to assign 4 arbitery bytes
- to an int. I was actually manipulating the bytes in MBR entry, I want
- to read the 4 bytes start from any address I specified(here I use the
- char[] to specify the address).
-
- So how could I get any 4 bytes in the MBR entry and convert it as an
- integer? Thank you again.
-
- Sincerely Ji

size = reinterpret_cast<int*>( somecharpointer );
*may* work for you, or it may not It depends on a lot on the architecture
you plan on running this on.

Some CPUs have problems reading integers that are not aligned on specific
byte boundaries, some do not. I believe (but could be mistaken) that AMD
and Intel are okay with this.

So, say, you had some char pointer pointing to arbitrary data that you can
read. If you are not worried about cross platform compatability, then I
would just point the int pointer to the start of where you think the array
is.

char* Data = SomeFunctionReturningChar*( someparm );
int* IntData = reinterpret_cast<int*>( Data );
At this point, if your architecture isn't too restrictive, you should be
able to read the contents of IntData as an integer, I.E.
std::cout << *IntData;
Should give you some int value, as long as the pointer points to memory you
have rights to read.
Notice, however, that incrementing your int pointer will increment it 4
bytes, not 1, because the compiler thinks it's int data you're pointing to.

++IntData;
will make IntData point to 4 bytes later, not one. Just as
IntData[1];
will look at the 5th through 8th bytes in the data.

Now, there are other ways to do it. There've been times I wanted to look at
data and didn't care to store it, so
std::cout << *reinterpret_cast<int*>( Data );
would give me an int value.
std::cout << *reinterpret_cast<int*>( Data + 1 );
would give me the 2nd through 5th bytes as an int, as would
std::cout << *reinterpret_cast<int*>( &Data[1] );

It really depends on what you are trying to accomplish how you would do it.
Looking and displaying the data the worst that usually can happen is you can
crash your program, reading data you don't own, etc.. Changing the data can
get you in trouble if you're not careful at what your'e pointing at, but
that's pretty much the same for all pointers.
Thank you very much for the detailed reply and abundant background
stated above.

Mar 17 '07 #14

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