c - unsigned int *代替int

我声明并定义一个函数，如下所示：

unsigned int doSomething(unsigned int *x, int y)
{
    if(1) //works
    if(y) //reports the error given below

    //I use any one of the ifs above, and not both at a time

    return ((*x) + y); //works fine when if(1) is used, not otherwise
}

我从main（）调用函数，如下所示：

unsigned int x = 10;
doSomething(&x, 1);

编译器报告错误和警告，如下所示：

passing argument 1 of 'doSomething' makes pointer from integer without a cast [enabled by default]|

note: expected 'unsigned int *' but argument is of type 'int'|

我尝试将所有可能的组合用于函数返回类型，函数调用以及参数类型。我哪里错了？

完整代码：

unsigned int addTwo(unsigned int *x, int y)
{
    if(y)
        return ((*x) + y);
}
int main()
{
    unsigned int operand = 10;
    printf("%u", addTwo(&operand, 1));
    return 0;
}

最佳答案

我也在Windows上使用了gcc 4.4.3。
该程序成功编译并产生输出“ 11”：

#include <stdio.h>

unsigned int doSomething(unsigned int *x, int y);

int main()
{
    unsigned int res = 0;
    unsigned int x = 10;

    res = doSomething(&x, 1);

    printf("Result: %d\n", res);

    return 0;
}

unsigned int doSomething(unsigned int *x, int y)
{
    if(y)
    {
        printf("y is ok\n");
    }

    if(1)
    {
        printf("1 is ok\n");
    }

    return ((*x) + y);
}

请检查是否适合您，然后将其与您的程序进行比较。
确保正确声明了该函数。

关于c - unsigned int *代替int，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/15064404/