无法确定输出C代码的原因

Question

I have the following code. I cannot determine the reason for the output it's giving.

char abc[14] = "C Programming";

printf("%s\n", abc+abc[3]-abc[4]);

output:rogramming

It is a quiz question with the following options: What would this print? a)C Programming b) rogamming c)Runtime Error d)Compilation Error

Answer 1

Perhaps it will be clearer if I put in stepping stones v3 and v4 . These are used to calculate an offset into the string of 3 indices. So the print output begins at abc[3] .

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

int main (int argc, char *argv[])
{
    char abc[14] = "C Programming";
    int v3 = abc[3];
    int v4 = abc[4];
    printf("%d - %d = %d\n", v3, v4, v3 - v4);
    printf("%s\n", abc + v3 - v4);
    return 0;
}

Program output:

114 - 111 = 3
rogramming

Answer 2

This may caused by the following: the value of abc[3] is 'r',the value of abc[4] is 'o'.so the value of

abc[3]-abc[4]

is 'r'-'o',change the value into int is 3 .So when you use printf("%s\\n", abc+abc[3]-abc[4]); ,it means you print this string from abc[3]. the following code will show you something.

 int main()
  {
  char abc[14] = "C Programming";
  printf("%d\n",abc[3]-abc[4]);
  printf("%s\n", abc+abc[3]-abc[4]);

  }

Answer 3

In this expression abc+abc[3]-abc[4] there is used so-called the pointer arithmetic.

An array name in expressions is converted to pointer to the first element of the array,

Thus in the expression above abc points to the first character of the string that is to 'C' abc + 1 points to space. abc + 2 points to 'P'. abc + 3 points to 'r' and so on.

expressions abc[3] and abc[4] are converted to type int and equal to internal codes of characters 'r' and 'o' The difference of these codes is equal to 3,

Thus expression abc+abc[3]-abc[4] can be written like abc + 3 and points to the forth element of the array that is to 'r' So the string starting from 'r' is outputed by statement

printf("%s\n", abc+abc[3]-abc[4]);

Answer 4

On virtually every platform (eg one using ASCII), the behaviour is undefined. So, none of the answers is correct.

abc+abc[3]-abc[4]

This expression is equal to

abc + 'r' - 'o'

the values of character constants are implementation-dependent; assuming an ASCII system, where r and o are 0x6f and 0x72, respectively, abc + 0x6f is out of the bounds of the array abc , leading to undefined behaviour. Note, that the later subtraction, appearing to bring the result back into bounds, doesn't change that; once reached, one cannot recover from undefined behaviour.

That said, the result is likely to be the same as abc + ('r' - 'o') , which, on an ASCII system, is abc + (0x6f - 0x72) , which in turn is abc + 3 and thus rogramming is outputted, as reported in the question.