string literal and pointer usage in calculating length of string

Question

The code below is related to calculating length of string

#include <stdio.h>

int strl (char *s);
main ()
{
  int len1 = str ("hah");
  printf ("%d", len1);
}


int
str (char *s)
{
  int length = 0;
  while (*s != '\0')
    {
      length++;
      s++;
    }

  return (length);
}

When I call the function str, I am assigning a string literal to pointer and not the address of the string literal to the pointer but it turn out that address is stored. How can pointers store address of "hah" without specifying its address?

we din metion adress of string to the pointer using sum &command — Fenris, Apr 07 '14 at 18:10
how come did it take the adress of "hah" just by assing it to the pointer — Fenris, Apr 07 '14 at 18:13
while assigning parameters to arguments we do s="hah"(s being a pointer) how can s store the base address of "hah" — Fenris, Apr 07 '14 at 18:16
I've formatted your code for you, and fixed some grammar, including turning the last sentence into a question. However, I'm still not sure what you are asking. — abligh, Apr 07 '14 at 18:17
appreciate ur patience have alook into comments of first answer — Fenris, Apr 07 '14 at 18:28

score 1 · Accepted Answer · edited May 23 '17 at 10:33

1

"hah" is of type char[] so when you pass it to function, it decays to pointer to its first element, as any other array.

edited May 23 '17 at 10:33

Community

1
1

answered Apr 07 '14 at 18:20

Dabo

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it is not any other array it is string literal and cant be edited as in case of arrays – Fenris Apr 07 '14 at 18:31
decays to point to its first element? how come does it do that without any operator – Fenris Apr 07 '14 at 18:33
because name of array is address of it's first element => `arr == &arr[0]` is always true. "string literal and cant be edited" it is not related to fact that name of array is address of its first elemnt – Dabo Apr 07 '14 at 18:39
1

How does it do that without any operator? It's a language design choice, and implementations are required to behave that way. – nobody Apr 07 '14 at 19:03
By the way, think what kind of alternative we have here ? To pass the array as is, value by value ? I think when `C` was designed it was very very expensive. – Dabo Apr 07 '14 at 19:12

score 0 · Answer 2 · edited Apr 07 '14 at 18:47

You don't need to be specifying its address, first of all - you've typed str("hah"); and not strl("hah"); (same when you wrote the function) - the string literal does actaully count as a variable in the function's stack memory and memory is allocated to it. What's more is that when appointing a string literal this way:

char* str = "hah";
char* sstr = "hah";

when printing:

printf("%d %d", str, sstr);

you'd get the same number. Using string literal - the compiler searches for the same string in the memory and if it's not there - it's allocating memory to it. In this case - there WAS a string the same so the addresses the pointers points to would be the same. The correct program would be this way:

#include <stdio.h>
int strl(char *s);
int main()
{ 
    int len1=strl("hah");
    printf("%d",len1);

    return 0;
}
int strl(char *s)
{
    int length=0;
    while (*s!='\0')
    {
        length++;
        s++;
    }

    return (length);
}

just changing str to strl as you mistook str for strl or the opposite :P

as you told that string literal is considered as variable .even in case of variables [suppose i to assign it to a pointer *p=&i ] — Fenris, Apr 07 '14 at 18:24
we used & operator which we din use while assing a string literal to a pointer — Fenris, Apr 07 '14 at 18:26

string literal and pointer usage in calculating length of string

2 Answers2