Using pointer declared in other function, in main.

**omnificient** · 07-03-2008

Hello,

I've got the following code:

Code:

#include <stdio.h>
void test();

int main()
{
	test();
	printf("%s",ptr);
	return 0;
}

void test()
{
	char *ptr;
	ptr = "Hello!";
}

Someone explained to me on IRC that string literals have static storage. So one would expect the preceding code to work. It however gives me a compiler error:

Code:

testcase.c: In function ‘main’:
testcase.c:7: error: ‘ptr’ undeclared (first use in this function)

Could somebody clarify this for me?

Thank You

**Elysia** · 07-03-2008

They do have static storage, but you aren't following scope rules.
The ptr "ptr" only exists within test.
A better example is to do this:

Code:

#include <stdio.h>
const char* test();

int main()
{
	printf("%s", test());
	return 0;
}

const char* test()
{
	const char* ptr = "Hello!";
	return ptr;
}

If this was a local array, it would not work, but for string literals, it will work.

**zacs7** · 07-03-2008

> If this was a local array, it would not work, but for string literals, it will work.
Yes, but it's a bad idea (at least I think so). The compiler can do what it wants really, "Hello!" might not even reside in read-only memory -- ie it could be moved out when test() returns.

I'd avoid it

Code:

const char * test(void)
{
    static const char str[] = "Hello!";
    return &str;
}

**Elysia** · 07-03-2008

I wouldn't place a function that returns a string literal in the first place

**omnificient** · 07-03-2008

The const keyword makes a variable unchangable. I don't see what it does to a function. Does it change the value returned by the function in a read only variable?

**Elysia** · 07-03-2008

A string literal is (mostly) read only, which is why I use const.
But const is const. The function returns const because the string literal is const. No biggie there.

**omnificient** · 07-03-2008

What I actually want to achieve is that I want to return a string to the function main, and print it there. In my book I read the following example:

Code:

#include <stdio.h>

void peasoup(int *green);

int main()
{
	int turn;
	int *head;
	
	turn=13;
	head = &turn;
	
	peasoup(head);
	
	printf("Oh no it's &#37;d\n", *head);
}

void peasoup(int *green)
{
	while((*green)--) // -- gaat voor de * er wordt dus 1 afgetrokken en vervolgens de geheugenlocatie bekeken
		puts("Blech!");
	
	*green = 13 * 51 + 3;
}

So I tried to apply it to strings.

**zacs7** · 07-03-2008

I think I could be wrong though, otherwise:

Code:

const char * blah(void)
{
    return "haha";
}

Wouldn't be right either. Perhaps someone smarter than us will come along

---------------

> So I tried to apply it to strings.

You could pass a pointer to a character array which you could then fill with the string...

Code:

void peasoup(char * str, size_t n)
{
    if(str == NULL)
        return;
    memset(str, 0, n);
    strncpy(str, "Pea soup!", n - 1);
    return;
}

/* ... */

char something[32];

peasoup(something, sizeof(something));

**nenpa8lo** · 07-03-2008

Originally Posted by omnificient

Hello,

Code:

testcase.c: In function ‘main’:
testcase.c:7: error: ‘ptr’ undeclared (first use in this function)

Simply main does not see 'ptr' and also it exists only for while when test() is called.

**Elysia** · 07-03-2008

Code:

#include <stdio.h>

void peasoup(int* green);

int main()
{
	int turn;
	turn=13;
	peasoup(&turn);
	printf("Oh no it's &#37;d\n", turn);
}

void peasoup(int* green)
{
	while((*green)--) // -- gaat voor de * er wordt dus 1 afgetrokken en vervolgens de geheugenlocatie bekeken
		puts("Blech!");
	
	*green = 13 * 51 + 3;
}

Cleaned.
No need to create a temporary pointer at all.

**omnificient** · 07-03-2008

I appreciate your help, but your code often contains concepts I don't understand. Why won't the following code work:

Code:

#include <stdio.h>
void test(char *green);

int main()
{
	char *green = "Hello\n";
	test(&green);
	printf("%s",green);
	return 0;
}

void test(char *green)
{
	*green = "Hi\n";
	return;
}

I get the following errors:

Code:

testcase.c: In function ‘main’:
testcase.c:7: warning: passing argument 1 of ‘test’ from incompatible pointer type
testcase.c: In function ‘test’:
testcase.c:14: warning: assignment makes integer from pointer without a cast

Can't I pass the function the memory adress and then let the function use that to modify the data? That sounds like a lot less complicated solution actually.

**Elysia** · 07-03-2008

First, I told you it should be const char*.
And then let me ask you two questions:
1) What happens when you dereference char*?
2) What is the type when you take the address of a variable?

If you can understand those concepts, you should able to understand why the code is wrong.

**The Dog** · 07-03-2008

You have two options:

1) Return a newly allocated pointer.
2) Pass a reference to the original pointer to the function. (So the function will accept a **)

Don't give up, struggle until you can't anymore... Then struggle some more

**C_ntua** · 07-03-2008

You are confusing the uses of * and & for pointers. If you have a variable the & in front of it gives you the adress. If you have a pointer then you already got the adrress.
So when a function needs a pointer to char as a parameter you either give a pointer or the address of a variable. Clear?

**Elysia** · 07-03-2008

Hey now, don't spoil the solution.
Actually, let me ask: what happens when you pass a pointer to a function that expects a pointer?
How is the contents of the pointer passed?

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