逗号作为分隔符和运算符 [英] Comma as a separator and operator

问题描述

所以我在某个地方遇到了这个问题:

So I came across this question somewhere:

情况1:

int a;
a = 1, 2, 3;
printf("%d", a);

情况2:

 int a = 1, 2, 3;
 printf("%d", a);

解释说:

第二种情况下会出错，因为使用逗号作为分隔符；在第一种情况下， = 优先于，，因此基本上是(a = 1)，2、3 ;

The second case gives error because comma is used as a separator, In first case = takes precedence over , so it is basically (a=1), 2, 3;

但是我想问为什么案例2中的 = 不优先于，?

But I want to ask why does = not take precedence over , in Case 2?

推荐答案

此

int a = 1, 2, 3;/* not a valid one */

是错误的，因为因为 = 具有更高的优先级，所以它在内部变为 int a = 1 ，并且没有 2 和 3 ，这就是为什么该语句无效并导致编译时错误的原因.

is wrong because since = has higher priority, so it become int a = 1 internally and there is no name for 2 and 3 thats why this statement is not valid and cause compile time error.

为避免这种情况，您可能要使用

To avoid this you might want to use

int a = (1, 2, 3); /* evaluate all expression inside () from L->R and assign right most expression to a i.e a=3*/

还有这里

int a;
a = 1,2,3; 

有两个运算符 = 和 ，请参阅 man运算符.赋值运算符 = 的优先级高于逗号运算符.因此它变为 a = 1 .

there are two operator = and , and see man operator. The assignment operator = has higher priority than comma operator. So it becomes a=1.

a = 1,2,3;
    | L--->R(coma operator associativity) 
    this got assigned to a

例如

int x = 10, y = 20,z;
z = 100,200,y=30,0; /* solve all expression form L to R, but finally it becomes z=100*/ 
printf("x = %d y = %d z = %d\n",x,y,z);/* x = 10, y = 30(not 20) z = 100 */
z = (100,200,y=30,0); /* solve all expression form L to R, but assign right most expression value to z*/ 

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