Explanation regarding character pointer for string

Question

Hi I'm new to programming. In the following code str is a pointer to a character, so str should contain the address of the character 'h'. Therefore %p should be used to print that address. But I don't understand how %s is used for printing a pointer parameter.

    #include<stdio.h>

    int main (){
    char s[] = "hello";
    char *str = s;
    int a[] = {1, 2, 3, 4, 5};
    int *b = a;
    printf("%s\n", str);        // I don't understand how this works ? 
    printf("%c\n", *str);       // This statement makes sense
    printf("%c\n", *(str + 1)); // This statement also makes sense.
    printf("%p\n",str);         // This prints the address of the pointer str. This too makes sense.
    printf("%d\n",*b);          // makes sense, is the same as the second print.
    //      printf("%d",b);     // I don't understand why str pointer works but this gives a compile error
    return 0;
}

What's not clear? Why a string format specifier works with a string? Because it is defined to work... — Eugene Sh.
– Eugene Sh., Commented May 18, 2016 at 21:48
I am not able to understand the first print. I think if the first print works the last print in the comment should also work but thats not true so thats my question, thanks. — op.dare7
– op.dare7, Commented May 18, 2016 at 21:49
@user6353278 What exactly you don't understand? How come str being an array hello is printed? — Iharob Al Asimi
– Iharob Al Asimi, Commented May 18, 2016 at 21:50
%s means: retrieve a char* argument. Dereference the pointer and print the character, and subsequent characters until a \0 is encountered. — Klas Lindbäck
– Klas Lindbäck, Commented May 19, 2016 at 9:56

lost_in_the_source · Accepted Answer · 2016-05-18 22:25:24Z

char s[] = "hello";

Declares an array of zero-terminated characters called s. Its the same as writing

char s[6] = { 'h', 'e', 'l', 'l', 'o', '\0' };

As you can see, the quotation marks are a shorthand.

char *str = s;

This declares str to be a pointer to a character. It then makes str point to the first character in s. In other words, str contains the address of the first character in s.

int a[] = {1, 2, 3, 4, 5};

Declares an array of integers. It initializes them to the values 1-5, inclusive.

int *b = a;

Declares b to be a pointer to an int. It then makes b point to the first int in a.

printf("%s\n", str);

The %s specifier accepts the address of the first character in the string. printf then walks from that address, printing the characters it sees, until it sees the \0 character at the end.

printf("%c\n", *str);

This prints the first character in str. Since str is pointing to a character (the first character in the string), then *str should obtain the character being pointed at (the first character in the string).

printf("%c\n", *(str + 1));

This prints the second character in str. This is the long way of writing str[1]. The logic behind this is pointer arithmetic. If str is the address of a character, then str + 1 is the address of the next character in the array. Since (str + 1) is an address, it may be dereferenced. Thus, the * obtains the character 1 character past the first character of the array.

printf("%p\n",str);

The %p specifier expects a pointer, just like %s would, but it does something else. Instead of printing the contents of a string, it simply prints the address the pointer is containing, in hex.

printf("%d\n",*b);

This prints the first int in the array pointed to by b. This is equivalent to writing b[0].

printf("%d",b);

b is an int *, not an int, which is what %d expects. If you were trying to print the address of the first element of the array, the specifier would be %p, not %d. Also, this line should not generate a compiler error. Instead, it should have been a runtime undefined behavior, since the compiler does not know what a printf format string is.

printf("%c\n", *str); prints the first character. *str is the simplified result of *(str + 0)); (where obviously + 0 does nothing). `

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