passing a pointer to a char array in c

Question

I wrote a small test program to see how I can pass a pointer to a char array to a function. This is my code:

#include <stdio.h>
#include <string.h>

int scopy(char *org ){
    printf("the value of org is %s" , org);
    printf("\n");
}

int main(void)
{
    char original[11];
    strcpy(original , "helloworld");

    scopy(original);

    printf("the value of original now is %s" , original);

    return 0;
}

My knowledge of pointers tells me that org is a pointer that is passed the memory address of original[0] , when I execute this program I observe a few things that has confused me and has caused me to doubt the facts I have learnt about pointers.

when I print org I get the complete word back . but according to the logic if pointers isnt this supposed to print out the memory address;

when I try printing out out[1] , out[2] , codepad tells me that the process has ended abnormally

My basic doubt in this question is what exactly is org doing , I understand it is a char pointer pointing to a memory address (which??) . Any help would be appreciated . I am very new to c programming .

Also when I try a while loop like

while(out[i] !='\0')
printf("%s",out[i]);

it does not terminate at the '\0' . the strcpy documentation states that the '\0' is copied to the char array.

teppic · Accepted Answer · 2013-03-04 23:46:59Z

3

org is a pointer to a char, which is what you're passing in scopy (a pointer to the first element of an array). When you use the specifier %s in printf you are treating a pointer to a char as a string; if you want to print its value as a pointer you should use %p instead. For example: printf("The memory address in org is %p\n", org);

From what you say I get the feeling you think that %s will convert a number to a string. It won't do that, it expects a pointer to the first character of a string.

Remember that the name of an array is equivalent to &array[0], i.e. when you pass the name of an array in a function you are passing a pointer to the first element. So here, when you call scopy(original) it's the same as writing scopy(&original[0]), and since original[0] is a char and & is a pointer to it, that's why your function takes char *.

edited Mar 4, 2013 at 23:46

answered Mar 4, 2013 at 22:59

teppic

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Comments

nos · Accepted Answer · 2013-03-04 23:00:14Z

Yes, you're right, when you call scopy(original); , scopy gets passed a pointer to the 1. element in the original array.

scopy(original); is the exact same as scopy(&original[0]);

However, you use %s with printf. %s expects the matching argument to be a char pointer , poiting to a sequence of chars, and the end of that sequence is a 0 byte. That's what C calls a string.

That means you can't do e.g. printf("%s",org[i]); , since org[i] is a char, it's not a char pointer into an array that ends with a 0 byte.

You can print a single char by doing prinf["%c", org[0]) `while(out[i] !='\0')

Or for your while loop:

while(out[i] !='\0')
  printf("%c",out[i]);

And you could also try:

while(out[i] !='\0')
  printf("%s",&out[i]);

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