我无法理解这段代码的工作原理。
#include<stdio.h>
void main(){
int a,b;
a=3,1;
b=(5,4);
printf("%d",a+b);
}
输出为7
。这个任务是什么?
答案 0 :(得分:4)
逗号运算符计算其第一个操作数并丢弃结果,然后计算第二个操作数并返回该值。
执行
后a=3,1; // (a = 3), 1;
a
将3
以及
b=(5,4); // Discard 5 and the value of the expression (5,4) will be 4
b
将4
。
Wikipedia上的更多示例:
// Examples: Descriptions: Values after line is evaluated:
int a=1, b=2, c=3, i=0; // commas act as separators in this line, not as an operator
// ... a=1, b=2, c=3, i=0
i = (a, b); // stores b into i
// ... a=1, b=2, c=3, i=2
i = a, b; // stores a into i. Equivalent to (i = a), b;
// ... a=1, b=2, c=3, i=1
i = (a += 2, a + b); // increases a by 2, then stores a+b = 3+2 into i
// ... a=3, b=2, c=3, i=5
i = a += 2, a + b; // increases a by 2, then stores a to i, and discards unused
// a + b rvalue. Equivalent to (i = (a += 2)), a + b;
// ... a=5, b=2, c=3, i=5
i = a, b, c; // stores a into i, discarding the unused b and c rvalues
// ... a=5, b=2, c=3, i=5
i = (a, b, c); // stores c into i, discarding the unused a and b rvalues
// ... a=5, b=2, c=3, i=3
return a=4, b=5, c=6; // returns 6, not 4, since comma operator sequence points
// following the keyword 'return' are considered a single
// expression evaluating to rvalue of final subexpression c=6
return 1, 2, 3; // returns 3, not 1, for same reason as previous example
return(1), 2, 3; // returns 3, not 1, still for same reason as above. This
// example works as it does because return is a keyword, not
// a function call. Even though most compilers will allow for
// the construct return(value), the parentheses are syntactic
// sugar that get stripped out without syntactic analysis
答案 1 :(得分:2)
a=3,1; // (a=3),1 -- value of expression is 1, side effect is changing a to 3
b=(5,4);
printf("%d",a+b); // 3 + 4
答案 2 :(得分:2)
在本声明中
a=3,1;
使用了两个运算符:赋值运算符和逗号运算符。赋值运算符的优先级大于逗号运算符的优先级,因此该语句等同于
( a = 3 ), 1;
1
被丢弃,因此为a
分配了值3
。
在本声明中
b=(5,4);
由于括号,首先评估逗号运算符。它的值是最后一个表达式4
的值。因此,为b
分配值4
。
结果,您得到的a + b => 3 + 4
等于7
。