c - printf() 函数并显示 %

Question

我正在尝试执行一个程序，该程序在 && 运算符返回 true 和返回 false 时打印数值。代码如下：-

#include <stdio.h>
main()
{
int a,b;
scanf("%d%d",&a,&b);
printf("Part I\n");
printf("(a%2 == 0) && (b%2 == 0): %d\n",(a%2 == 0) && (b%2 == 0));
printf("(a%3 == 0) && (b%3 == 0): %d\n",(a%3 == 0) && (b%3 == 0));
printf("(a%5 == 0) && (b%5 == 0): %d\n",(a%5 == 0) && (b%5 == 0));
printf("(a%7 == 0) && (b%7 == 0): %d\n",(a%7 == 0) && (b%7 == 0));
printf("Part II\n");
printf("The AND operator yields: %d\n",(a%2 == 0) && (b%2 == 0));
printf("The AND operator yields: %d\n",(a%3 == 0) && (b%3 == 0));
printf("The AND operator yields: %d\n",(a%5 == 0) && (b%5 == 0));
printf("The AND operator yields: %d\n",(a%7 == 0) && (b%7 == 0));

return 0;
}

输出（连同我的输入）如下： -

210
210
Part I
(a%2 == 0) && (b%2 == 0): %d
(a%2 == 0) && (b%2 == 0): %d
(a%2 == 0) && (b%2 == 0): %d
(a%2 == 0) && (b%2 == 0): %d
Part II
The AND operator yields: 1
The AND operator yields: 1
The AND operator yields: 1
The AND operator yields: 1

为什么第一部分以这种方式表现？即使我将 && 替换为 || 也会发生这种情况。我正在使用 Borland C++ 编译器 5.5 。请帮忙。

Answer 1

因为如果你想实际显示一个%，那么你必须在printf格式字符串中用另一个转义它%。例如

printf("(a%%2 == 0) && (b%%2 == 0): %d\n",(a%2 == 0) && (b%2 == 0));
          ^              ^

Answer 2

我已经用http://codepad.org/对此进行了测试，我认为它使用了 gcc，并且代码运行正常。%但是您可能会尝试在文字%（即）之前添加额外的内容，%%以便编译器知道后面的 % 是实际字符。像这样：

printf("Part I\n");
printf("(a%%2 == 0) && (b%%2 == 0): %d\n",(a%2 == 0) && (b%2 == 0));
printf("(a%%3 == 0) && (b%%3 == 0): %d\n",(a%3 == 0) && (b%3 == 0));
printf("(a%%5 == 0) && (b%%5 == 0): %d\n",(a%5 == 0) && (b%5 == 0));
printf("(a%%7 == 0) && (b%%7 == 0): %d\n",(a%7 == 0) && (b%7 == 0));

Answer 3

您实际上是%在第一部分使用非法转义序列字符进行打印。这就是 printf 产生垃圾值的原因。

printf("(a%2 == 0) && (b%2 == 0): %d\n",(a%2 == 0) && (b%2 == 0));
              ^ ^
      这是你误会了

它应该像

printf("(a%%2 == 0) && (b%%2 == 0): %d\n",(a%2 == 0) && (b%2 == 0));  

您还可以阅读C 中使用的所有格式说明符。