Adding an integer to C-String while printing

Question

  int main(){
printf("hello world"+2);

}

test.c:32:25: warning: adding 'int' to a string does not append to the string
      [-Wstring-plus-int]
    printf("hello world"+2);
           ~~~~~~~~~~~~~^~
test.c:32:25: note: use array indexing to silence this warning
    printf("hello world"+2);
                        ^
           &            [ ]
1 warning generated.
alpha@Desktop % ./a.out   
llo world%    

So this is what I am getting. And if I increase the number then it is simply slicing the string by that.Can anyone explain this output and why it's happening to me?

Answer 1

In the C and C++ languages, every string that has been defined with double quotes is of type char* (pointer to char) as a statically allocated char array. When you call printf() on such string it does something like this:

void printf(char* str) {
    while(*str!= '\0') {
        // write the current character to stdout
        write(STDOUT_FILENO, str, sizeof(char));
        str ++;
    }
}

I don't know if it really is written that way, indeed it probably isn't, the functionality is the same. It iterates over the string until it finds the null-terminator ( '\\0' ). All pointers are almost equivalent to long and when you add a number to them, it increments their underlying value with sizeof(type) , where type is the type of value the pointer points towards. Thus, when you add a number to "Hello World", makes printf() think your string starts on a different memory address and prints 'llo World'.

If you want to print the string with the value '2' appended to the end of it, as @Elliott Frisch stated, you would use print("Hello World%d", 2) .

I would suggest looking at sprintf() and strcat() for string concatenation in C.

Answer 2

In C, what we might think of as a string is actually an array of characters. What you pass to printf() in both cases is a pointer to the start of such an array. You can do arithmetic with pointers, which is what you are doing here.

Here's some clarification.

#include <stdio.h>

int main(void) {
    // A
    printf("Hello, world!\n");
    // B
    printf("Hello, world!\n" + 2);
}

Output:

Hello, world!
llo, world!

In case A , you point to the start of an array containing the H character, and it reads until the last one (which is in fact a 'this is the end'-character: \\0 ).

⬇️
H	e	l	l	o	,		w	o	r	l	d	!	\\n	\\0

In case B , you add 2 to the pointer before passing it to printf() . Now printf() starts reading here:

		⬇️
H	e	l	l	o	,		w	o	r	l	d	!	\\n	\\0

This is what causes your 'slice'.

Answer 3

The rules of the C language do not allow you to do anything like this. Your question could be compared to the question why I pour milk into the tank of the car instead of gasoline, and it does not go. The computer is just a stupid piece of hardware and it can only do what it was taught.