String problems

[PhilipRoe]

If you try to "cout" the address of an array of characters you get the array itself.
So how do you get the address if you really want it? (There are a few answers on the web like static casts to <void> and <void>* but they don't seem to work.)
One time we need to know this is when debugging a program which claims we've done an illegal operation.


Another thing

I don't like using the <string> library because I don't like using code I don't understand. Anyway why should Bill Gates have all the fun? So instead of

count = strlen(myString);

I write

char* p =myString;
while(*p) p++;
count=p-mystring;

but this doesn't work if myString has been declared with

const char* mystring = 0;

because the compiler cannot convert const char* to char*. This strikes me as absurd. Like saylng you can't have x=3 because 3 is constant.

So if myString is a const pointer I have to grit my teeth and use strlen and strcpy. (And of course strcpy requires the source to be const.

Grateful for help

[emaghero]

If you try to "cout" the address of an array of characters you get the array itself.
So how do you get the address if you really want it? (There are a few answers on the web like static casts to <void> and <void>* but they don't seem to work.)
One time we need to know this is when debugging a program which claims we've done an illegal operation.


Another thing

I don't like using the <string> library because I don't like using code I don't understand. Anyway why should Bill Gates have all the fun? So instead of

count = strlen(myString);

I write

char* p =myString;
while(*p) p++;
count=p-mystring;

but this doesn't work if myString has been declared with

const char* mystring = 0;

because the compiler cannot convert const char* to char*. This strikes me as absurd. Like saylng you can't have x=3 because 3 is constant.

So if myString is a const pointer I have to grit my teeth and use strlen and strcpy. (And of course strcpy requires the source to be const.

Grateful for help

If you want to output the address of an array you should use the reference operator &.

double *ptr;
cout<<"The address of the pointer is "<<&ptr<<endl;
 

Also Bill Gates isn't responsible for the <string> library. That was written long before Microsoft existed. So you're not using anything Microsoft specific if you use it. Just be sure that you adhere to the ANSII standard.

The reason libraries such as <string> exist is because they help programmers to save time when using certain types and functions in C++. The functions contained in such libraries are very well written and are completely general. They are there so that you don't have to reinvent the wheel when it comes to very common functions like strcpy and strlen. If these libraries didn't exist then every programmer would have his/ her own version of strlen, which would not make for very efficient code, especially if you intend using your code on different compilers.

[Strika Amaru]

...how do you get the address if you really want it?

Haven't worked with cout for a while, and furthermore i wouldn't recommend it for debugging because it's buffered and if/when a program dies, the output dies with it in the buffer.
Instead, you could use (from <stdio.h>) printf( <string template>, <unlimited number of arguments> ). Sound strange? First time i've seen it, in full detail, i ran back to cout. So I won't go into details :) To see a pointer, the code looks something like this:

SomeType *ptr;
 // ....
printf( "%d\n", ptr );

"%d\n" is the 'template' of the output; it says, "Print a decimal number and a new line". "ptr" is the number in question. That's about it.

Just thought about this: in cout, do you write *ptr, or ptr? If it's the first, then the poor function is working properly...

char* p =myString;
while(*p) p++;
count=p-mystring;

but this doesn't work if myString has been declared with

const char* mystring = 0;

because the compiler cannot convert const char* to char*. This strikes me as absurd. Like saylng you can't have x=3 because 3 is constant.

So if myString is a const pointer I have to grit my teeth and use strlen and strcpy. (And of course strcpy requires the source to be const.

Several points here:
1. strcpy doesn't *require* the source to be const; const in an argument list is an indication to the compiler; it says "this function doesn't alter this parameter, so it's ok to call it for a constant." If you have a function, say, void foo( char* param ), and you try to pass it a const char*, the compiler will protest; it is forced to maintain the integrity of any const, so it won't accept it being passed to an "unsafe" function.
2. const char* mystring = 0; This one seems bizarre. I just went in gcc and it works without any problems, both with const char* and char*. Are you declaring p as const? Because that's all I can think you doing wrong. const char* p says: "the POINTER is constant". Essentially, you're not assigning x=3, you're assigning 3=4, which is definitely not kosher.

I don't like using the <string> library because I don't like using code I don't understand. Anyway why should Bill Gates have all the fun?

Because library functions are years old, they are definitely correct. And usually they are optimized to the umpteenth level, too...

[weaknessforcats]

Haven't worked with cout for a while, and furthermore i wouldn't recommend it for debugging because it's buffered and if/when a program dies, the output dies with it in the buffer.
Instead, you could use (from <stdio.h>) printf( <string template>, <unlimited number of arguments> ).

I recommend that you never debug ussiong display statements in your code. Learn to use your debugger. The problem is that those displays need to be taken out, which changes the source code and that not requires a complete new test pass.

Further, in C++ avoid printf() and its lik like vsprintf(), sprint(), etc. They only work woth the built-in types from C. Use an C++ inserter instead.

[AdrianH]

I’m sorry to say that there are quite a few errors in this thread, but I don’t have time to go into them all, so I will just answer the question.

If you try to "cout" the address of an array of characters you get the array itself.
So how do you get the address if you really want it? (There are a few answers on the web like static casts to <void> and <void>* but they don't seem to work.)

cout << static_cast<void*>(ptr); is what you are looking for.

One time we need to know this is when debugging a program which claims we've done an illegal operation.


Another thing

I don't like using the <string> library because I don't like using code I don't understand. Anyway why should Bill Gates have all the fun? So instead of

count = strlen(myString);

I write

char* p =myString;
while(*p) p++;
count=p-mystring;
but this doesn't work if myString has been declared with

const char* mystring = 0;

Bill has nothing to do with this. As stated upthread, it is to keep you from having to reinvent the wheel. However, to fix your code you would do this:

char const * p =myString;  // Problem is on this line (put const in front of char if you wish)
while(*p) p++;
count=p-mystring;
 

This will still allow you to increment the pointer but will also let you assign a const char * to it. Note, this code may not be as efficient as the code provided by the compiler.

As a side note, I put const after the thing I am consting. This is because it is consistent. Putting const in front of the entire expression is the only time that const refers to the type after const.

because the compiler cannot convert const char* to char*. This strikes me as absurd. Like saylng you can't have x=3 because 3 is constant.

No, it’s more like saying 3=x. You cannot assign a constant a new value.

So if myString is a const pointer I have to grit my teeth and use strlen and strcpy. (And of course strcpy requires the source to be const.

No, strcpy() doesn’t require that the source be const, the prototype is saying that when passed, the function strcpy() will not modify the source.

Careful, you grit your teeth any harder and they may shatter. ;)

Grateful for help

Glad to give it.


Adrian

P.S. Using printf is a viable alternative, but you should use either %p (pointer address) or if you want to print as a hex address of a consistent length use %08x and cast the pointer to int. Using %d as stated also works but I find that it isn’t as neat because the length can be very variable. Also, casting is a good idea because there are compilers out there that will check the parameters after the format string against the format string to ensure you are not doing anything wrong.