passing pointer to functions in C

Question

Go straightforward, since C pass pointers as parameters to functions, why the program below the printf in swap function doesn't print the same address as the pinrtf in main function(I think the pointers were passed correctly), does something wrong here?

#include <stdio.h>

void swap(char **str1, char **str2)
{
        char * temp = *str1;

        *str1 = *str2;
        *str2 = temp;

        printf("1---(%#x) (%#x)---\n", &str1, &str2);
        printf("2---(%s) (%s)---\n", *str1, *str2);
}


int main ()
{

        char * str1 = "this is 1";
        char * str2 = "this is 2";
//      swap(&str1, &str2);     

        printf("(%s) (%s)\n", str1, str2);
        printf("(%#x) (%#x)\n", &str1, &str2);
        swap(&str1, &str2);
        printf("(%s) (%s)\n", str1, str2);
        printf("(%#x) (%#x)\n", &str1, &str2);

        return 0;
}

C uses pass by value; `str1` and `str2` inside the function are local variables to that function, and so have different addresses to variables in any other function — M.M, Feb 12 '16 at 03:34
Maybe you meant the first printf to print `str1` and `str2` rather than `&str1` and `&str2` — M.M, Feb 12 '16 at 03:35
Why use `"%#x"` to print a pinter rather than standard `"%p"`? — chux - Reinstate Monica, Feb 12 '16 at 03:35
A question similar to this is answered in detail here: http://stackoverflow.com/questions/8403447/swapping-pointers-in-c-char-int — Turk, Feb 12 '16 at 03:36
You have six different `printf` calls that print a total of 12 values. Can you explain which values you don't think should be different from which other values. — David Schwartz, Feb 12 '16 at 03:37
Thank you guys for the help, I am a newbie of C programming, so the code that I posted maybe a little tricky here, and I think I the answers that I looked for, here is a link http://denniskubes.com/2012/08/20/is-c-pass-by-value-or-reference/ Thank you again guys !! — Fulei Tang, Feb 26 '16 at 17:02

score 2 · Answer 1 · answered Feb 12 '16 at 04:13

Here in swap function

swap(&str1, &str2);

You send the swap function, address of str1 & str2. In the swap function

void swap(char **str1, char **str2)

You create a variable to save the address (which is have their own address).

With your print function

printf("1---(%#x) (%#x)---\n", &str1, &str2);

Here you print the address of the variable that store the address of your char. If you print what is stored in that address, you can find your char address. To print what is stored, just used reguler print like this

printf("0---(%#x) (%#x)---\n", str1, str2);

After you run it, you get something like this

(this is 1) (this is 2)
(0x61fedc) (0x61fed8)
0---(0x61fedc) (0x61fed8)---
1---(0x61fec0) (0x61fec4)---Here is the address of var that store your char address
2---(this is 2) (this is 1)---
(this is 2) (this is 1)
(0x61fedc) (0x61fed8)

score 1 · Answer 2 · answered Feb 12 '16 at 03:45

In the swap function you are trying to print the address of the local variables str1 and st2 which is different from the address of the variables str1 and str2 in the main function. Try chnaging the print statement in swap function to:

 printf("1---(%#x) (%#x)---\n", str1, str2);

score 0 · Accepted Answer · answered Feb 26 '16 at 17:03

Thank you guys for the help, I am a newbie of C programming, so the code that I posted maybe a little tricky here, and I think I the answers that I looked for, here is a link http://denniskubes.com/2012/08/20/is-c-pass-by-value-or-reference/ Thank you again guys !!

passing pointer to functions in C

3 Answers3