printf语句错误？(在C中)*更新*

Question

为了澄清目的,我需要程序打印输入a和b的数字,而不是实际的字母a和b.
好的,这是每个yall建议的修订程序:

int main (int argc, char *argv[])  
{
   int a; /*first number input*/  
   int b; /*second number input*/  

   a = atoi(argv[1]); /*assign to a*/  
   b = atoi(argv[2]); /*assign to b*/  

   if (a < b)  
      printf("%s\n", a < b); /* a is less than b*/  
      else {  
         printf("%s\n", a >= b); /* a is greater than or equal to b*/  
      }  

   if (a == b)  
      printf("%s\n", a == b);  /* a is equal to b*/  
      else {  
         printf("%s\n", a != b); /* a is not equal to b*/  
      }  

   return 0;  
} /* end function main*/  

大声笑,现在当我运行程序时,我被告知

8 [main] a 2336 _cygtls::handle_exceptions: Error while dumping state   
Segmentation fault 

这是什么意思？(如果你现在还没有注意到我对这个东西很无望lol).

Answer 1

您要求printf()打印布尔表达式的值(对于true和false,它们总是分解为1或0).

您可能希望您的代码看起来更像:

if (a < b)
     printf("%s\n", "a < b"); /* a is less than b*/
else {
     printf("%s\n", "a >= b"); /* a is greater than or equal to b*/
}

将结果显示为字符串.

Answer 2

这一行:

if (a = b)

不应该

if (a == b) 

同样在这里:

printf("%d\n", a = b);  /* a is equal to b*/

应该

printf("%d\n", a == b);  /* a is equal to b*/