#include <stdio.h>
int main()
{
int feet,fathoms;
fathoms = 2;
feet = 6*fathoms;
printf("There are %d feet in %d fathoms!\n",6*fathoms);
printf("Yes,I said %d feet!\n",6*fathoms);
return 0;
}
为什么我第七行第二个%d输出是6356868
printf("There are %d feet in %d fathoms!\n",6*fathoms);
->
printf("There are %d feet in %d fathoms!\n",6*fathoms, fathoms);
或者
printf("There are %d feet in %d fathoms!\n",feet, fathoms);
【以下回答由 GPT 生成】
问题优化:
问题标题: C语言编程中遇到第七行输出错误的问题
问题内容:
#include <stdio.h>
int main()
{
int feet,fathoms;
fathoms = 2;
feet = 6*fathoms;
printf("There are %d feet in %d fathoms!\n",6*fathoms);
printf("Yes, I said %d feet!\n",6*fathoms);
return 0;
}
问题: 在上述C语言代码中,为什么第7行的第二个%d
输出结果是6356868
?
回答:
第7行的第二个%d
输出结果是6356868
是因为在printf
函数的第一个参数中,占位符和参数的数量不匹配,导致结果错误。
具体解决方案如下:
原始代码中第7行的printf
语句如下:
printf("There are %d feet in %d fathoms!\n",6*fathoms);
在这里,%d
需要两个参数来替代占位符,但实际上只有一个参数6*fathoms
。占位符和参数数量不匹配会导致未定义行为,具体输出结果是不确定的。
要解决这个问题,可以将第7行修改为:
printf("There are %d feet in %d fathoms!\n",6*fathoms, fathoms);
这样,第一个%d
将被替换为6*fathoms
的值,而第二个%d
将被替换为fathoms
的值。
修复后的代码如下:
#include <stdio.h>
int main()
{
int feet,fathoms;
fathoms = 2;
feet = 6*fathoms;
printf("There are %d feet in %d fathoms!\n",6*fathoms, fathoms);
printf("Yes, I said %d feet!\n",6*fathoms);
return 0;
}
这样修改后,第7行的输出结果将是There are 12 feet in 2 fathoms!
,正确显示了占位符的值。
【相关推荐】