scanf函数声明 [英] Scanf statement

问题描述

 的#include＆LT;＆stdio.h中GT;
＃包括LT＆;＆stdlib.h中GT;
 

您好，我想知道为什么这个系统并不执行程序。它好像我一直具有正常履行scanf函数功能的麻烦。我得到的错误：

错误：与属性warn_unused_result声明scanf函数的忽略返回值

请注意：我知道有写我目前的方案更有效的方法，但这是题外话。我想知道这是什么错误意味着，为什么我不断收到它，可能的解决方案。感谢你们。这里是code：

  INT主（INT ARGC，CHAR *的argv []）{    INT X;
    诠释Ÿ;
    INT Z者除外;
    INT温度;    scanf函数（％d个，＆安培; X）;
    scanf函数（％d个，＆安培; Y）;
    scanf函数（％d个，＆安培; Z）;        如果（X＆GT; Y）{
           TEMP = X;
           X = Y;
           Y =温度;
        }        如果（Y＆GT; Z）{
            温度= Y;
            Y = Z;
            Z =温度;
        }        如果（X将Z）{
            TEMP = X;
            X = Z者除外;
            Z =温度;
        }        的printf（％D，X）;
        的printf（％D，Y）;
        的printf（％D，Z）;  返回EXIT_SUCCESS; }
 

解决方案

从GCC文档：

  

的 warn_unused_result 属性会导致发出警告
       如果函数与此属性的调用者不利用其
       返回值。这是地方为函数不检查有用
       结果要么是一个安全问题，或者总是一个bug，如
       的realloc 。

  INT FN（）__attribute__（（warn_unused_result））;
      INT富（）
      {
        如果（FN（）℃，）返回-1;
        FN（）;
        返回0;
      }
 

在警告结果第5行。

#include <stdio.h>
#include <stdlib.h>

Hello, I was wondering why this program doesn't execute. It seems as though I've always been having trouble with performing the scanf function properly. I get the error of:

error: ignoring return value of scanf declared with attribute warn_unused_result

Note: I know there is a more efficient method for writing my current program, but that is beside the point. I want to know what this error means and why I keep getting it and possibly a solution. Thank you guys. Here is the code:

int main (int argc, char *argv[]){

    int x;
    int y;
    int z;
    int temp;

    scanf("%d", &x);
    scanf("%d", &y);
    scanf("%d", &z);

        if (x > y) {
           temp = x;
           x = y;
           y = temp;
        }

        if (y > z) {
            temp = y;
            y = z;
            z = temp;
        }

        if (x > z) {
            temp = x;
            x = z;
            z = temp;
        }

        printf("%d", x);
        printf("%d", y);
        printf("%d", z);

  return EXIT_SUCCESS; }

解决方案

From the GCC documentation:

The warn_unused_result attribute causes a warning to be emitted if a caller of the function with this attribute does not use its return value. This is useful for functions where not checking the result is either a security problem or always a bug, such as realloc.

      int fn () __attribute__ ((warn_unused_result));
      int foo ()
      {
        if (fn () < 0) return -1;
        fn ();
        return 0;
      }

results in warning on line 5.

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