Using pointer declared in other function, in main.

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

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.

> 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;
}

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

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?

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.

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.

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));

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.

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.

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.

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.

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

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?

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?