为什么这段代码输出什么？

Question

Consider this "EXAM" question:

int main()
{
   int a=10,b=20;
   char x=1,y=0;
   if(a,b,x,y)
   {
      printf("EXAM");
   }
}

Please let me know why this doesn't print anything at all.

Answer 1

Comma operator - evaluates 1st expression and returns the second one. So a,b,x,y will return value stored in y, that is 0.

Answer 2

a,b,x,y是y （因为逗号运算符计算其右操作数的结果），y为0，这是假的。

Answer 3

The comma operator returns the last statement, which is y . Since y is zero, the if-statement evaluates to false, so the printf is never executed.

Answer 4

Because expression a,b,x,y evaluates to y , which in turn evaluates to 0 , so corresponding block is never executed. Comma operator executes every statement and returns value of last one. If you want logical conjunction, use && operator:

if (a && b && x && y) { ... }

Answer 5

Others already mentioned that the comma operator returns the right-most value. If you want to get the value printed if ANY of these variables is true use the logical or:

int main()
{
   int a=10,b=20;
   char x=1,y=0;
   if(a || b || x || y)
   {
      printf("EXAM");
   }
 }

But then be aware of the fact that the comma evaluates all expressions whereas the or operator stops as soon as a value is true. So with

int a = 1;
int b;
if(a || b = 1) { ... }

b has an undefined value whereas with

int a = 1;
int b;
if(a, b = 1) { ... }

b will be set to 1.