运算符优先级 - 示例 [英] Operator Precedence - sample

问题描述

 main（）{
  int  i，j，k; 
 k =（i =  4 ，j =  5 ）; 
 printf（  k =％d，k）; 
  return   0 ; 
} 

哪个获得了优先权以及如何获得？

解决方案

< blockquote>输出 k = 5 因为：

1. 编译告诉我。
2. 这是怎么回事逗号运算符工作，即：[逗号]是一个二进制运算符，它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（和类型）（参见维基百科的Comma运营商 [ ^ ]）。
因此评估步骤为：


1. i = 4 （丢弃）
2. j = 5 （已退回）
3. k = 5 （ k 获取逗号运算符的返回值）

转到此链接，你会知道 pr-order ecedence-in-c-programming-language [ ^ ]

main(){
     int i,j,k;
     k = (i = 4, j = 5);
     printf("k = %d", k);
     return 0;
}

which gets the prcedence and how ?

解决方案

output is k = 5 because:

1. Compiler told me.
2. This is how comma operator works, namely: "[comma] is a binary operator that evaluates its first operand and discards the result, and then evaluates the second operand and returns this value (and type)" (see Comma operator at Wikipedia[^]).
Hence evaluation steps are:

1. i = 4 (discarded)
2. j = 5 (returned)
3. k = 5 (k gets the returned value of the comma operator)

Go to this link and you will know the order-of-precedence-in-c-programming-language[^]

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