忽略属性warn_unused_result [-Wunused-result]声明的'int scanf（const char *，...）'的返回值？ [英] ignoring return value of ‘int scanf(const char*, ...)’, declared with attribute warn_unused_result [-Wunused-result]?

问题描述

当我编译如下程序时：
g ++ -O2 -s -static 2.cpp 它给了我警告忽略返回值'int scanf（const char *，...）'，声明属性为warn_unused_result [-Wunused-result] 。

但是当我删除 -02 来自复制语句不显示警告。

我的 2.cpp 程式：

 ＃include > 
 int main（）
 {
 int a，b; 
 scanf（％d％d，& a，& b）; 
 printf（％d \ n，a + b）; 
返回0; 
} 
  

这个警告的含义是什么？是 -O2的含义 ??

解决方案

这意味着你不要检查scanf的返回值。

它可能很好地返回1（仅设置a）或0（不设置a和b）。

编译时未显示优化时未显示的原因是，除非启用优化，否则需要分析才能看到此内容。 -O2启用优化 - http://gcc.gnu.org/onlinedocs/gcc /Optimize-Options.html 。

只要检查返回值，就会删除警告，并使程序以可预测的方式运行，如果它没有收到两个数字：

  if（scanf（％d％d，& a，& b）！= 2）
 {
 //做某事，比如.. 
 fprintf（stderr，期望至少有两个数字作为输入\ n）; 
 exit（1）; 
} 
  

When I compiled the following program like: g++ -O2 -s -static 2.cpp it gave me the warning ignoring return value of ‘int scanf(const char*, ...)’, declared with attribute warn_unused_result [-Wunused-result].
But when I remove -02 from copiling statement no warning is shown.

My 2.cpp program:

#include<stdio.h>
int main()
{
   int a,b;
   scanf("%d%d",&a,&b);
   printf("%d\n",a+b);
   return 0;
}

What is the meaning of this warning and what is the meaning of -O2 ??

解决方案

It means that you do not check the return value of scanf.

It might very well return 1 (only a is set) or 0 (neither a nor b is set).

The reason that it is not shown when compiled without optimization is that the analytics needed to see this is not done unless optimization is enabled. -O2 enables the optimizations - http://gcc.gnu.org/onlinedocs/gcc/Optimize-Options.html.

Simply checking the return value will remove the warning and make the program behave in a predicable way if it does not receive two numbers:

if( scanf( "%d%d", &a, &b ) != 2 ) 
{
   // do something, like..
   fprintf( stderr, "Expected at least two numbers as input\n");
   exit(1);
}

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