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return a string

Hello,

How to I do to return a string as a result of a function.
I wrote the following function:

char prt_tralha(int num)
{
int i;
char tralha[num];

tralha = "#";
for( i = 0; i < num-1; i++ )
strcpy(tralha, strcat(tralha," #"));

return tralha;
}

And I really like to use it, thus:

int main()
{
printf("%s \n", prt_tralha(5));

return 0;
}

But when I compile it, gcc shows this message:

tmp.c: In function `prt_tralha':
tmp.c:16: error: incompatible types in assignment
tmp.c:20: warning: return makes integer from pointer without a cast
tmp.c:20: warning: function returns address of local variable

Thanks,

Nascimento
Nov 14 '05 #1
23 3567
Nascimento wrote:
Hello,

How to I do to return a string as a result of a function.
I wrote the following function:

char prt_tralha(int num)
{
int i;
char tralha[num];

tralha = "#";
for( i = 0; i < num-1; i++ )
strcpy(tralha, strcat(tralha," #"));

return tralha;
}

And I really like to use it, thus:

int main()
{
printf("%s \n", prt_tralha(5));

return 0;
}

But when I compile it, gcc shows this message:

tmp.c: In function `prt_tralha':
tmp.c:16: error: incompatible types in assignment
tmp.c:20: warning: return makes integer from pointer without a cast
tmp.c:20: warning: function returns address of local variable

Thanks,

Nascimento


#include <stdio.h>
#include <stdlib.h>
const char * string_function (void)
{
static const char * my_text = "My text";
return my_text;
}

int main(void)
{
printf("%s\n", string_function ());
return EXIT_SUCCESS;
}

Fundamentally, you pass a pointer to a string; not
the string.

--
Thomas Matthews

C++ newsgroup welcome message:
http://www.slack.net/~shiva/welcome.txt
C++ Faq: http://www.parashift.com/c++-faq-lite
C Faq: http://www.eskimo.com/~scs/c-faq/top.html
alt.comp.lang.l earn.c-c++ faq:
http://www.comeaucomputing.com/learn/faq/
Other sites:
http://www.josuttis.com -- C++ STL Library book
http://www.sgi.com/tech/stl -- Standard Template Library
Nov 14 '05 #2

Nascimento wrote:
Hello,

How to I do to return a string as a result of a function.
I wrote the following function:

There are two ways to "return" a string from a function. One is to
simply return it...
char prt_tralha(int num)
I.E. char * ptr_tralha(int num)

Here we return a pointer to char, or string. The problem with this is
the return value can't be an automatic variable, so you'll need to work
with malloc.

Another method is to make the string part of the argument list. This is
probably prefered, since the memory management takes place outside of
the function and is (arguably) easier.

I.E. char ptr_tralha(int num, char *ret)
{
int i;
char tralha[num];
Assuming the later method, tralha would simply be the argument to this
function, and not declared here.

tralha = "#";
for( i = 0; i < num-1; i++ )
strcpy(tralha, strcat(tralha," #"));

return tralha;
There would be no need to "return" tralha here, since it was passed in
via pointer in the argument list.
}

And I really like to use it, thus:

int main()
{
printf("%s \n", prt_tralha(5));
try this instead:

char tralha[5];
ptr_tralha(5,tr alha);
printf("%s \n", tralha);

return 0;
}

But when I compile it, gcc shows this message:

tmp.c: In function `prt_tralha':
tmp.c:16: error: incompatible types in assignment
Well, of course. The function is (was) returning a char, but you were
trying to use it as if it were a char *.
tmp.c:20: warning: return makes integer from pointer without a cast
tmp.c:20: warning: function returns address of local variable

Thanks,
Hope this helps. I'm sure others will have suggestions as well.

-Jason
Nascimento


Nov 14 '05 #3
Nascimento,
you cannot assign a char to a string like that:
tralha = "#"
you can do it like this: tralha[0]='#';
A more elegant code for your purpose goes like this:

--- tralha.c ---

void
ptr_tralha(char *s)
{
while (*s != '\0') /* while the contents of s are different
* from NULL */
*s++ = '#'; /* s = '#' . increment s */
}
int
main(void)
{
char s[5];
ptr_tralha(s);
puts(s);

return 0;
}

Nov 14 '05 #4

Nascimento wrote:
int i;
char tralha[num];

tralha = "#";
for( i = 0; i < num-1; i++ )
strcpy(tralha, strcat(tralha," #"));


Oh, no! Look what's happening here! You are trying to copy chars into
the tralha[] array without using the proper notation. This is bad.

What you want to do is use array substript notation or pointer
arithmetic. Substript is by far easier for the beginner.

tralha[0] = '#'; /* Use single quotes for char */
for( i = 1; i < num-1; i++ ) /* i=1, not 0 (see above) */
tralha[i] = '#'; /* again, copy a char only */
tralha[i] = '\0'; /* don't forget the terminator! */

You don't need (or want) strcpy or strcat when you are building an
array from scratch.

Nov 14 '05 #5
un************@ gmail.com (Nascimento) writes:
How to I do to return a string as a result of a function.
I wrote the following function:

char prt_tralha(int num)
{
int i;
char tralha[num];

tralha = "#";
for( i = 0; i < num-1; i++ )
strcpy(tralha, strcat(tralha," #"));

return tralha;
}

And I really like to use it, thus:

int main()
{
printf("%s \n", prt_tralha(5));

return 0;
}

But when I compile it, gcc shows this message:

tmp.c: In function `prt_tralha':
tmp.c:16: error: incompatible types in assignment
tmp.c:20: warning: return makes integer from pointer without a cast
tmp.c:20: warning: function returns address of local variable
All three error messages are correct.

tralha = "#";
error: incompatible types in assignment

You can't assign strings like that. The name of a string variable (or
of any array) is implicitly converted, in most contexts, to a pointer
to its first element. See section 6 of the C FAQ,
<http://www.eskimo.com/~scs/C-faq/top.html>. You can initialize a
char array with a string literal, but I don't think that's what you
want in this case.

return tralha;
warning: return makes integer from pointer without a cast
warning: function returns address of local variable

You declared your function to return a char (a single character
value). You're trying to return a char*. The types are incompatible.
(The "integer" in the first warning refers to type char, which is an
integer type.)

Even if prt_tralha() were declared to return a char*, returning the
address of a local variable would be invalid. Your array tralha
ceases to exist as soon as the function terminates. Your main program
would receive a pointer to a non-existent object, and any attempt to
use it will invoke undefined behavior. In the worst case, it will
work as expected, failing only at the most inconvenient possible
moment. (The compiler isn't required to diagnose this error, but gcc
is kind enough to do so anyway.)

Other problems:

Your declaration "char tralha[num];" declares a variable length array
(VLA). This is a new feature in C99. If you're not concerned about
portability to compilers that don't support VLAs, that's fine, but you
should be aware that such support is not (yet?) universal. You might
consider using malloc() to allocate the memory dynamically.
for( i = 0; i < num-1; i++ )
strcpy(tralha, strcat(tralha," #"));


What is the purpose of the strcpy()? strcat() appends "#" to your
string (assuming tralha is a valid string in the first place) and
returns a pointer to its first element. The strcpy() copies the
nstring onto itself. That's either undefined behavior or a no-op, I
don't remember which.

What you're trying to do is set tralha to a string of num '#'
characters. Using strcat() for this is wasteful.

for (i = 0; i < num; i ++) {
tralha[i] = '#';
}
tralha[num] = '\0';

or

memset(tralha, '#', num);
tralha[num] = '\0';

And since you need space for the trailing '\0', the size of tralha had
better be at least num+1.

Since C doesn't treat arrays as first-class objects (you can't assign
them, compare them, or pass them as parameters, at least not
directly), dealing with character strings can be tricky, especially
when you don't know until execution time how big they're going to be.

If you want a function to return a variable-sized string (or any
array) to its caller, there are basically 3 ways to do it.

1. Let the caller allocate the array. The caller then needs to pass
in the address of the array and its size (and you need to decide what
to do if the caller's array isn't big enough). See the standard
fgets() function for an example.

2. Return a pointer to a static variable. Since static variables
continue to exist after the function returns, this avoids the problem
you had. The drawback is that there's only one allocated result; if
you call your function multiple times, each call will clobber the
previous result. This is especially bad in the presence of recursion
<OT>or multi-threading</OT>.

3. Allocate the result array in the function using malloc(). This is
probably the most flexible method, but it then require the caller to
free() the array when it's finished with it.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.
Nov 14 '05 #6


Nascimento wrote:
Hello,

How to I do to return a string as a result of a function.
I wrote the following function:

char prt_tralha(int num)
{
int i;
char tralha[num];

tralha = "#";
for( i = 0; i < num-1; i++ )
strcpy(tralha, strcat(tralha," #"));

return tralha;
}

And I really like to use it, thus:

int main()
{
printf("%s \n", prt_tralha(5));

return 0;
}

But when I compile it, gcc shows this message:

tmp.c: In function `prt_tralha':
tmp.c:16: error: incompatible types in assignment
tmp.c:20: warning: return makes integer from pointer without a cast
tmp.c:20: warning: function returns address of local variable


You seem to be confused about several points. The
comp.lang.c Frequently Asked Questions (FAQ) list

http://www.eskimo.com/~scs/C-faq/top.html

covers some of them: Section 6 will help you understand
gcc's first complaint, Question 8.1 explains the second,
and Question 7.5 covers the third.

However, gcc did not catch all your errors. One that
it didn't catch is the strcpy() call: you are passing it
two strings that overlap, but strcpy() requires the source
and destination areas to be distinct. When you try to use
strcpy() on overlapping strings, anything at all might
happen. (Since the overlap is "perfect" -- the source and
destination are exactly the same string -- the strcpy()
call is pointless anyhow. I think this goes back to your
misunderstandin gs about strings; see FAQ section 8.)

Read the FAQ, re-read your textbook, and start over.
Good luck!

--
Er*********@sun .com

Nov 14 '05 #7
On 29 Apr 2005 14:04:10 -0700, un************@ gmail.com (Nascimento)
wrote:
Hello,

How to I do to return a string as a result of a function.
I wrote the following function:

char prt_tralha(int num)
Here, you're saying to return a character, not a string. You want
char * prt_tralha(int num)
where the returned value will be a pointer to a character array
(string).
{
int i;
char tralha[num];
You can't declare a variable size array. What you can do is
char *tralha;
tralha = malloc(num);
but remember to check the return value to make sure the malloc
succeeded. One possibility is
if (tralha == NULL)
return NULL;
Also, since you are allocating new memory, it needs to be freed at
some point. More below on this.
tralha = "#";
You can't just assign strings. (Your first compiler error below.) You
can do
strcpy(tralha, "#");
for( i = 0; i < num-1; i++ )
strcpy(tralha, strcat(tralha," #"));
This makes little sense, but I think I can guess what you're trying to
do. Make a string with num '#'s, right? How about
for( i = 0; i < num-1; i++ )
strcat(tralha, "#");
return tralha;
This is OK now, since you've allocated new memory for tralha. In your
original, you returned the address of a local array, which will be
invalid as soon as you return from the function. This is why you got
the third compiler error. The second error was because you were
returning a pointer, when you told the compiler you wanted to return a
char.}

And I really like to use it, thus:

int main()
{
printf("%s \n", prt_tralha(5));
Now, we have the problem that prt_tralha has allocated memory which
needs to be freed. In your test program it doesn't matter, but in a
real program, every call to prt_tralha would allocate more memory, and
you could eventually run out. That's what we call a memory leak. One
way to fix this is

char *temp = prt_tralha(5);
if (temp != NULL)
{
printf("%s \n", temp);
free temp;
}
return 0;
}

But when I compile it, gcc shows this message:
Some hints: Look up the definitions and try to find example uses of
every library function you use, such as strcpy, strcat, malloc, free
above. I referred to all your compiler diagnostics as "errors" on
purpose. Unless you know exactly why the compiler issued a warning,
and you know for certain that it's harmless, consider it an error and
fix it. There is hardly ever a good reason for ignoring compiler
warnings.

Get a good tutorial. I recommend "The C Programming Language" by
Kernighan and Ritchie, which is a good tutorial and reference.
tmp.c: In function `prt_tralha':
tmp.c:16: error: incompatible types in assignment
tmp.c:20: warning: return makes integer from pointer without a cast
tmp.c:20: warning: function returns address of local variable

Thanks,

Nascimento


--
Al Balmer
Balmer Consulting
re************* ***********@att .net
Nov 14 '05 #8
mu********@yaho o.com-dot-br.no-spam.invalid (iru_muzgo) writes:
Nascimento,
you cannot assign a char to a string like that:
tralha = "#"
you can do it like this: tralha[0]='#';
A more elegant code for your purpose goes like this:

--- tralha.c ---

void
ptr_tralha(char *s)
{
while (*s != '\0') /* while the contents of s are different
* from NULL */
No. NULL is (a macro that expands to) a null *pointer* constant.
'\0' is a null character, sometimes referred to as NUL. Using the
term NULL to refer to a character value is misleading.
*s++ = '#'; /* s = '#' . increment s */
}
int
main(void)
{
char s[5];
ptr_tralha(s);
puts(s);

return 0;
}


Your array s is not initialized before you pass its address to
ptr_tralha(). Inside ptr_tralha(), you loop over the array until you
find a '\0' character, but there's no reason to assume that you ever
will. Unless there happens to be a '\0' character somewhere within s,
the loop in ptr_tralha() will go past the end of the array, invoking
undefined behavior.

I just tried compiling and running your program, and it printed a
string of 5 '#' characters followed by a newline. Apparently there
just happened to be no '\0' characters in s itself, but there just
happened to be a '\0' character immediately following it in memory.
This is just one of the infinitely many possible consequences of
undefined behavior.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.
Nov 14 '05 #9
Alan Balmer <al******@att.n et> writes:
On 29 Apr 2005 14:04:10 -0700, un************@ gmail.com (Nascimento)
wrote:

[...]
{
int i;
char tralha[num];


You can't declare a variable size array.


Yes you can, if you have a C99 compiler or a pre-C99 compiler that
supports the feature. (The form of the diagnostics implies that he's
probably using gcc, which does support VLAs.)

I covered the portability issues in my previous followup.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.
Nov 14 '05 #10

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