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1)how (s[i]!=0) evaluates?
2)how does *(s+i) works?
3) how does i[s] works?
please help me with these douts

What I have tried:

#include<stdio.h>
#include<conio.h>
int main()
{
    char s[]="peak";
    int i=0;
    while(s[i]!=0)
    {
        printf("%c%c \n",s[i],*(s+i));
        printf("%c%c \n",i[s],*(i+s));
        i++;
    }
    return 0;
}
Posted
Updated 24-Jun-22 12:36pm
v2

1 solution

C doesn't have any concept of strings past "string literal": a sequence of characters enclosed in double quotes.

So when you are traversing a char array, you need to know either how long the string is, or look for a string terminator character. C chose the latter: and selected the character '\0' as the terminator.
This is a null value which evaluates to zero when you try to use it in any form of math, including comparisons.
So s[i] != 0 is looking for the terminating character: the end of the string.

In C, the name of an array is defined as a pointer to the first element in the array, so *(s + i) adds the value of i to the address of the first element, and accesses that element: it's the equivalent of s[i]

i[s] doesn't even compile ... it's meaningless garbage.
 
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Comments
k5054 24-Jun-22 23:40pm    
"i[s] doesn't even compile ... it's meaningless garbage."
Not true. In C and C++ 5[x] is perfectly valid, as is explained here:
https://stackoverflow.com/a/381549

I should add that except for lecture notes, trying it for yourself, obfuscated programming contests, and just someone being a right git, you should never see this usage in real life.

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