While loop without Condition

Question

Following is the program I encountered

#include <stdio.h>       

void sp_to_dash(const char *str);
int main(void)
{
    sp_to_dash("this is a test");
    return 0;
}
void sp_to_dash(const char *str)
{
    while(*str) //beginning of while loop
    {               
        if(*str== ' ')printf("%c", '-');
        else printf("%c", *str);
        str++;
    }
}

At the beginning of while loop we can see that while(*str) is initiated without any condition, which means it is not mentioned that when the *str should stop on null or '\0'. The code is working fine, however according to me it should be like this while(*str != null) or while(*str != '\0'). Please explain this one to me.

Answer 1

It is same as writing this -

while (*str != '\0')

'\0' is of type int and is equivalent to 0 and will not effect execution of loop.

Note- Don't compare string with == and there is no operator !==.

This will cause error if you use this (comparing char with string literal) -

while (*str != "\0")  

Answer 2

The while loop needs a value as a condition in order to work. When you insert a condition like while (*str == '3'), your condition checks if *str actually is '3' and if so it is interpreted as 1 (you can think of this as True if you like) and if not, it interprets it as 0 (or False).

This is why you can write while (0) or while (1).

Now, your condition is the value of *str so, during the condition check, you will check the value of *str to see if it is 0 or not (every other value (-1, 7, 32333.34 ...) considered as True) str is a const char * so we need to translate char value into int values (using ASCII). The char that has the value 0 is '\0'.

Answer 3

It is exectly the same condition as (*str!='\0') because the ascii value of \0 is 0 so the loop will stop when *str will be \0, means at the end of the string,

*any other value except 0 is a true condition for itself (like *str) and will cause the while loop to continue e,g: while(5){} will be an endless loop.