C progarmming

Question

#include <stdio.h>
int main(void) {
    int i = 0;
    char ch;
    while(1) {
        ch= *("hello world" + i);
        if(ch == '\0')
            break;
        putchar(ch);
        i++;
    }
    return 0;
}

What does  ch= *("hello world" + i); do?

Comments

*("hello world") means it is pointing to first char of "hello world"

What I mean to say, let char p[] = "hello world", then *p menas we are pointing to first char of hello world.

Same thing is applied here,

instead of writing *(p+i), they wrote directly *("hello world" + i)

So, ch = *("hello world" + 0) => ch ='h'

ch = *("hello world" + 1) => ch = 'e' and so on...

It will print the whole string hello world.

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Is the string "hello world" automatically stored temporarily somewhere by the compiler every-time that statement is executed? Coz the terminating condition is \0. How does compiler know \0 is there after "hello world"? \0 is added when some char array is storing it. But this is not the case here.

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hii @gauravkc

 How does compiler know \0 is there after "hello world"? \0 is added when some char array is storing it.

the bolded part you said is not correct, string written inside double inverted commas(" ") are known to compiler as a string constant, and has an implicit termination with '\0'.

for eg: at compile time when compiler sees "hello", it will mark it as a string constant and provide one implicit pointer 'p' pointing to it. now whenever you will write/refer "hello", you will be actually accessing 'p'

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okk.. Thanks :)