无效的二进制操作数/(具有"int *"和"int")? [英] Invalid Operands to binary / (have 'int *' and 'int')?

问题描述

每次我尝试以下方法:

long crypt(int *integer)
{
printf("Enter five digit integer:\n");  
scanf("%i",integer);

int digit1=integer/10000;
int digit2=(integer%10000)/1000;
int digit3=(integer%1000)/100;
int digit4=(integer%100)/10;
int digit5=(integer%10)/1;

const char *digit1c[10];
const char *digit2c[10];
const char *digit3c[10];
const char *digit4c[10];
const char *digit5c[10];

(还有更多，但这似乎是问题所在，我将根据要求添加其余内容.)

(There's more but this seems to be the problem, I'll add the rest by request.)

然后返回此错误:

math2.h:44:20: error: invalid operands to binary / (have ‘int *’ and ‘int’)
math2.h:45:21: error: invalid operands to binary % (have ‘int *’ and ‘int’)
math2.h:46:21: error: invalid operands to binary % (have ‘int *’ and ‘int’)
math2.h:47:21: error: invalid operands to binary % (have ‘int *’ and ‘int’)
math2.h:48:21: error: invalid operands to binary % (have ‘int *’ and ‘int’)

我知道它与我用来初始化数字的运算符有关，我确实尝试将其类型更改为"int *"，但这没有用. 那么，这里到底发生了什么?

I know it has something to do with the operators I used to initialize the digits and I did try changing their type to "int *" but that didn't work. So what's happening here exactly?

推荐答案

integer是int(int*)的指针，因此，当您要使用int时，它会指向到，您需要取消引用:

integer is a pointer to int (int*), so when you want to use the int it points to, you need to dereference it:

int digit1=(*integer)/10000; // and so on...

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