[英]Order of execution of C operators
#include <stdio.h>
int main(void) {
int i;
i = 10;
i*= 10+2;
printf("%d",i);
return 0;
}
why is the output of the following code 120 and not 102? 为什么下面的代码120而不是102的输出?
Because the order of precedence makes '+' higher than *=, so the 10+2 will occur befor the i *=. 因为优先顺序使'+'高于* =,所以10 + 2将在i * =之前出现。
C reference for ordering at http://en.cppreference.com/w/c/language/operator_precedence 有关订购的C参考,网址为http://en.cppreference.com/w/c/language/operator_precedence
In this Line i*= 10+2; 在这一行中,i * = 10 + 2; In this case the 12 multiply by i It Means i=10*12;
在这种情况下12乘以i表示i = 10 * 12; So it Will Give 120 In answer
所以它会给120的答案
To solve this issue Try This. 解决此问题的方法请尝试此。
i*= 10;
i+=2;
your code work like . 您的代码像。
i= i*(10+2)
so it give the answer like 120. 所以给出的答案是120。
if you wanna answer like 102 the do . 如果你想像102那样做。
i=i*10+2
This 这个
i*= 10 + 2;
is syntactic sugar for 是语法糖
i= i * (10 + 2);
the rest is precedence left to right, add and subst after mult./division 其余的优先级从左到右,在多/除后加和减
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