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Here is a program, I write it to output all the characters of a string one by one. But I also print the address of individual blocks of the array. The problem is addresses for all blocks are same. Why? Does someone know?

#include<stdio.h>
int main()
{
    char enter[]="Kinsman";
    char *ptr;  
    ptr=enter;
    int i=0;
    while(*ptr!='\0')
    {
        printf("%c%p\n",*ptr,&ptr);
        ptr++;
        for(i=0;i<=100000000;i++);
    }
return 0;
}
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8
  • 5
    for(i=0;i<=100000000;i++); ?? why this Commented Oct 30, 2013 at 10:50
  • 1
    @GrijeshChauhan Back on my old ATARI 800XL, waiting one second was accomplished easily with FOR I=1 TO 500: NEXT I. But this method - waiting in terms of cycles - has been outdated for about 20 years. Commented Oct 30, 2013 at 11:17
  • @glglgl Yes I read it somewhere in past and it will consume CPU cycles. ~~ But a C compiler can optimize it by replacing i = 100000000;.. it is not reliable. Correct? Commented Oct 30, 2013 at 11:22
  • @GrijeshChauhan I used it just to give a pause for each output through loop. Commented Oct 30, 2013 at 11:28
  • 1
    @AshishTomer It is better to use the appropriate function - sleep() or usleep() on Unix, Sleep() on WIndows. Commented Oct 30, 2013 at 11:36

3 Answers 3

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4

Because you print the address of the actual pointer.

When you use &ptr you get the address of the actual pointer and not the address if points to. Remove the ampersand (address-of operator &) so you have only ptr.

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7 Comments

Yes you are right. Thank you! :) Can you tell me how do I clear screen on startup using clrscr()?
@AshishTomer By calling the function?
No i know that actually I added conio.h but GCC gives error: "reverse.c:2:18: fatal error: conio.h: No such file or directory #include<conio.h> ^ compilation terminated."
@AshishTomer What is the point of clearing the "screen"? As a user, this would annoy me as I lose track of what happened before. Is this really absolutely necessary?
@glglgl no! It's not necessary. I just wanted to :P :D
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3

You are printing the address of the pointer, not the value of the pointer

Try

printf("%c%p\n",*ptr, static_cast<void*>(ptr));

(https://stackoverflow.com/a/18929285/259)

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7 Comments

I printed the value and address both using %c and %p
@AshishTomer Yes but you also added &, read answer again. &ptr should be just ptr in printf
@AshishTomer: You might want to see this post for clrscr.
Using %p without a void * is undefined behavior. Unless the argument is a ptr-to-void, a cast is required.
@DavidSykes what are standards? And where can I get them?
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1

ptr is a pointer and it is also a variable in stack that has an address. This is fixed, while what it points to is varying by ptr++, thus you've to print the pointed-to value and not the address of the pointer itself.

 printf("%c%p\n",*ptr, (void*)ptr);
 //                   ^  remove & , and add void*
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1 Comment

Can you tell me how do I clear screen? <br/>I added conio.h but GCC gives error: "reverse.c:2:18: fatal error: conio.h: No such file or directory #include<conio.h> ^ compilation terminated."

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