Question

我正在尝试创建一个程序，它将输出16个数字的二进制代码。以下是我到目前为止的情况：

#include <stdio.h>
#include <stdlib.h>
int i;
int count;
int mask;

int i = 0xF5A2;
int mask = 0x8000;
int main()
{
printf("Hex Value= %x Binary= \n", i);
{
    for (count=0; count<15; count++1)
    {
        if (i&mask)
            printf("1\n");
        else
            printf("0\n");
    }
    (mask = mask>>1);
}
return 0;
}

错误：

|16|error: expected ')' before numeric constant|

如果我有任何其他错误，也请告诉我，提前致谢！

Answer 1

错误是指这个表达式：

count++1

这没有任何意义。

我假设你想要：

count++

制作专栏

for (count=0; count<15; count++)

您的代码中还有其他陌生感，例如：

int i;              // Declare an integer named "i"
int mask;           // Declare an integer named "mask"

int i = 0xF5A2;     // Declare another integer also named "i".  Did you forget about the first one???
int mask = 0x8000;  // Did you forget you already declared an integer named "mask"?

printf("Hex Value= %x Binary= \n", i);
{
    [...]
}  // Why did you put a bracket-scope under a PRINTF call?
   // Scopes typically follow loops and if-statements!

 (mask = mask>>1);  // Why put parens around a plain-old expression??

<小时/> 在修复代码中的怪异之后，它应该如下所示：

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int i = 0xF5A2;
    int mask = 0x8000;

    printf("Hex Value= %x Binary= \n", i);
    for (int count=0; count<15; ++count, mask>>=1)
    {
        printf("%d\n", (i&mask)? 1 : 0);
    }
    return 0;
}

无法找到导致此错误的原因

1 个答案: