可以请解释一下这个程序的 output 是如何浮动的

Question

can some please explain how float is the output of this program I am unable to understand

#include <stdio.h>
int main()
{
    int x = 1;
    short int i = 2;
    float f = 3;
    if (sizeof((x == 2) ? f : i) == sizeof(float))
        printf("float\n");
    else if (sizeof((x == 2) ? f : i) == sizeof(short int))
        printf("short int\n");
}

Answer 1

Indeed the ternary operator is not returning quite what you think it is, but it is also not quite what @user15790236 thinks either. The ternary operator is an operator, and as such it expects arguments of the same type on either side of the colon. Since you have different types, the integer is promoted to float, and thus the expression is of type float.

Answer 2

The ternary operator is not returning quite what you think it is.

( (x == 2)? f: i ) actually returns the value of i, which is by value and is then handled as an int, not a short int - thus, 4 bytes in size according to your 64bit compiler. Put another way, the return from the ternary operator is a copy of what is in i, not i itself.

Prove this to yourself by checking the value of:

sizeof(2)

To get what you seem to expect, you could instead say:

( (x==2) ? sizeof(f) : sizeof(i) )

The next question you may want to consider is why does the compiler pick int for the type when a short would do the trick. Have a peek at Implicit type promotion rules

Answer 3

Since C11, to note the type , rather attempt sizeof(type) compares, use _Generic .

#include <stdio.h>

#define type(X) _Generic((X), \
    short: "short", \
    int: "int", \
    float: "float", \
    double: "double", \
    default: "default" \
)

int main(void) {
  int x = 1;
  short int i = 2;
  float f = 3;
  puts(type(x));
  puts(type(i));
  puts(type(f));
  puts(type((x == 2) ? f : i));
}

Output

int
short
float
float

(x == 2)? f: i (x == 2)? f: i is a float as @SGeorgiades well points out, the type of the a?b:c is the same, regardless if a is true or not.