c - 请向我解释为什么这个带有 if 语句的简单 C 程序不起作用？

Question

我刚刚了解了 if 语句并尝试制作某种计算器，但它不起作用。它要求您输入一个运算(目前只能进行加法)，然后要求输入两个整数。 super 简单，但是行不通。这个错误对你们来说可能是显而易见的，但我就是看不到。请帮忙!代码如下:

int main()
{
int operation;
int addition;
float firstNumber; 
float secondNumber;
printf("Type in an operation.\n");
scanf(" %s", operation);


if(operation = addition){
printf("Please, enter an integer.\n");
scanf(" %f", &firstNumber);

printf("Please, enter a second integer.\n");
scanf(" %f", &secondNumber);

printf("Answer: %d", firstNumber + secondNumber);
}else{
printf("Sorry, only addition works..");
}
return 0;
}

Answer 1

为什么你的代码应该有效？

int operation; 
/* int for storing a string? huh? 
 * or were you thinking about function pointers ?
 * Normally you would use a char[]
 */


scanf(" %s", operation); 
/* Using %s specifier looks weird .
 * Also scanf is reading into operation and not &operation
 * So is having a space in the in the beginning of the format string.
 */

int addition; 
 /* Automatic variables are not initialized as per the standard
  * But what about the type?
  * Were you intending to do something like char addition[]="addition"
  */

if(operation = addition)  
/* If you somehow manage to get to this point 
 * you have another problem you do an assignment in the statement using =
 * You should have been  using ==
 */