Problem with char arrays

**JOCAAN** · 12-31-2010

Hi,
I have a problem with char arrays and pointers. I think it's an obvious begginer issue but I can't figure out why do I get this.
I have this piece of code in my program:

Code:

                char readbuffers[3] = "bbb";
		char readbuffer[3] = "aaa";
		printf("%s\n", readbuffer);


		char *codCliente = argv[2];
		char *tipoCliente = argv[3];
		char *Distancia = argv[7];
		char *tipoTrayecto = argv[8];
		printf("%s\n", codCliente);	
		printf("%s\n", tipoCliente);
		printf("%s\n", Distancia);
		printf("%s\n", tipoTrayecto);

And my output is a seven characters array, aaabbb��

Why the arrays are automatically appending?

Thank you in advance.

**tabstop** · 12-31-2010

"bbb" is a four-character string. So is "aaa". When you try to stuff that four-character string into a three-character array, you lose the fourth character: the "end-of-string" marker \0, which means you no longer have a string (so trying to pretend it is a string, for instance by printing it with %s, is doomed to failure).

**JOCAAN** · 12-31-2010

Oh, thank you very much tabstop. That's right. I was stuck so much time in this silly thing!

**kona49er** · 12-31-2010

Originally Posted by JOCAAN

Oh, thank you very much tabstop. That's right. I was stuck so much time in this silly thing!

To let the compiler do the counting you can use:

Code:

        char readbuffer[] = "abc";

this also works:

Code:

        char *readbuffer = "abc";

**vart** · 01-01-2011

Originally Posted by kona49er

this also works:

Code:

        char *readbuffer = "abc";

in this case you should declare it as const - since you do not allocate the memory and just have a pointer to the constant string that cannot be modified.

Code:

        const char *readbuffer = "abc";

and in the case it is technically not a buffer - so the name would be misguiding

**JOCAAN** · 01-02-2011

Thank you.

**kona49er** · 01-02-2011

[QUOTE=vart;992964]in this case you should declare it as const - since you do not allocate the memory and just have a pointer to the constant string that cannot be modified.

Code:

        const char *readbuffer = "abc";

Your statement, above, is incorrect.

Storage in memory must be allocated for it. The difference between:

Code:

        char *a = "abc";

and

Code:

        const char *a ="def";

the later declares that the storage allocated is not intended to be modified and
and a error should be displayed when the program attempts to make a change
to any element of this array.

with the former, array elements may be modified with impunity.

**laserlight** · 01-02-2011

Originally Posted by kona49er

Storage in memory must be allocated for it.

Good catch, but...

Originally Posted by kona49er

the later declares that the storage allocated is not intended to be modified and
and a error should be displayed when the program attempts to make a change
to any element of this array.

I think vart's point is that storage is not allocated for an array named readbuffer as readbuffer would merely be a pointer to the first element of a string literal.

Originally Posted by kona49er

with the former, array elements may be modified with impunity.

The point is, the storage allocated is not intended to be modified either way. Whether you declare that the pointer points to a const char or not does not change this as attempting to modify any element of the string literal results in undefined behaviour. It is not generally true that these "array elements may be modified with impunity" if you declare readbuffer as a pointer to non-const char.

**kona49er** · 01-02-2011

Originally Posted by laserlight

Good catch, but...

It is not generally true that these "array elements may be modified with impunity" if you declare readbuffer as a pointer to non-const char.

Yes, my bad. It does work with gcc but that's not a guarantee.

**tabstop** · 01-02-2011

Originally Posted by kona49er

Yes, my bad. It does work with gcc but that's not a guarantee.

It compiles everywhere, gcc or no, since there is no syntax error involved. It runs nowhere (unless you have a system that does not protect string literals, but I am unaware of such).

**quzah** · 01-03-2011

Originally Posted by tabstop

It compiles everywhere, gcc or no, since there is no syntax error involved. It runs nowhere (unless you have a system that does not protect string literals, but I am unaware of such).

Borland's Turbo C, I believe, lets you modify string literals. I think Adak keeps bringing that point up.

Quzah.

**kona49er** · 01-03-2011

Originally Posted by tabstop

It compiles everywhere, gcc or no, since there is no syntax error involved. It runs nowhere (unless you have a system that does not protect string literals, but I am unaware of such).

The following code runs fine on cygwin compiled with gcc.

Code:

#include <stdio.h>

int main()
{
        char * rb = "hij";

        printf("%s \n", rb);
        rb[0] = 'z';
        printf("%s \n", rb);
}

But ... due to the enlightenment received from this thread, I will not write the code this way.

**anduril462** · 01-03-2011

There are a great number of systems that allow modifying string literals. Many embedded systems don't have a memory management unit that allows memory segmentation, thus you can't mark a segment read-only to protect string literal data. Often in these systems, all memory, including the text section, is writable. Makes debugging pointer problems a nightmare, as there is no seg fault generated when there is no segment or segment rule to violate, so there's nothing for a debugger to trap.

You also, however, have the issue that certain compilers can/may place string literals in writable memory segments (Turbo C as Quzah/Adak point out). Older versions of gcc had the -fwritable-strings flag that did this, but they dropped it somewhere between version 2.9 and 4.1, realizing it was a bad idea (we had tons of code that relied on that "feature" at my last job. When we upgraded gcc to 4.1, it was a nightmare).

String literals should never be counted on as being modifiable. If you want to modify the contents of a string literal, you are really asking the compiler to let the data vary at run time, which means the data in the string literal (not the address thereof) should be in a variable, like char[] = "foo". Besides, not all compilers support writable strings, and you don't have a good mechanism for determining the size of the modifiable string you have to work with. Also, guaranteeing that a string literal can't change allows the compiler to optimize for size by having all string literals you use that have identical contents live at the same address, so "abc" exists only once, now matter how many things you make point to it.

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