Difference between *str and atoi(str)

I was tokenizing, and used strtok on a text file (which has been read into an array 'sto开发者_如何学JAVAre') with the delimiter '='

so there was a statement in the file : TCP.port = 180

And I did:

str = strtok(store, "=");

str= strtok(NULL, "=");

Now if I do *str, it gives me '82' (probably some junk value) but atoi(str); gives me 180 (the correct value)

I was hoping someone could shed light onto this, shouldn't dereferencing str give me 180 too?

Compile and run this program. It should give you a better idea of what's going on.

#include <stdlib.h>
#include <stdio.h>

int main(void) {
    const char *s = "180";
    printf("s       = \"%s\"\n", s);
    printf("atoi(s) = %d\n", atoi(s));
    printf("*s      = %d = '%c'\n", *s, *s);
    return 0;
}

Here's the output:

s       = "180"
atoi(s) = 180
*s      = 49 = '1'

No. atoi gives you the integer represented by the string str points to. Dereferencing str (*str) gives you the value of the char str points to (which is not the value you wrote).

You need to understand exactly how strings work in C to see what's going on here. The str variable is a pointer to the first character in the string. Dereferencing str gives the value pointed to by str, namely the character '1'. Similarly, dereferencing str+1 will give you the next character, '8'. You can see how the end of the string is signified with *(str+3) (or, equivalently, str[3]), which gives a null byte. The function atoi knows how to interpret the characters as a base-10 string of ASCII characters, which is much more complicated than a dereference.