Question

int main(int argc, const char * argv[])
{

    int age = 40;
    float gpa = 3.25f;
    char grade ='A';
    double fun = 2.000043f;
    char companyName[20] = "O'Brien Enterprises";

    int *pAge = &age;
    int *pGpa = &gpa;
    int *pGrade = &grade;
    int *pFun = &fun;
    int *pCompanyName = &companyName;

    printf("Value of variables through pointers:\n");
    printf("age = %i\n", *pAge);
    printf("gpa = %f\n",*pGpa);
    printf("grade = %c\n", *pGrade);
    printf("fun = %d\n", *pFun);
    printf("companyName = %s\n", *pCompanyName);


    return 0;
}

当我运行此代码时，Xcode会回复大量警告和错误。在声明和初始化除age之外的所有变量的指针时，它会显示incompatible pointer types。在尝试打印出来时，除了age，format specifies a different type以外的所有内容都会显示。为什么是这样？当代码按原样运行时，我得到以下结果：

Value of variables through pointers:
age = 40
gpa = 0.000000
grade = A
fun = -2147483648
(lldb)

Answer 1

您已将所有这些声明为整数的指针，而它们应该是指向int，float，char的指针，等等！

int *pAge = &age;
float *pGpa = &gpa;
char *pGrade = &grade;
double *pFun = &fun;
char **pCompanyName = &companyName;

Answer 2

这是正确的代码

#include <stdio.h>

int main(void) 
{
    int age = 40;
    float gpa = 3.25f;
    char grade ='A';
    double fun = 2.000043f;
    char companyName[20] = "O'Brien Enterprises";

    int *pAge = &age;
    float *pGpa = &gpa;
    char *pGrade = &grade;
    double *pFun = &fun;
    char *pCompanyName = companyName;

    printf("Value of variables through pointers:\n");
    printf("age = %i\n", *pAge);
    printf("gpa = %f\n",*pGpa);
    printf("grade = %c\n", *pGrade);
    printf("fun = %f\n", *pFun);
    printf("companyName = %s\n", pCompanyName );


    return 0;
}

输出

Value of variables through pointers:
age = 40
gpa = 3.250000
grade = A
fun = 2.000043
companyName = O'Brien Enterprises

您必须在printf函数中使用corect指针类型和格式说明符。

或者，如果您想要一个指向字符数组的指针，那么您可以编写

char ( *pCompanyName )[20] = &companyName;

并相应地

printf("companyName = %s\n", *pCompanyName );

通过解除引用指针获取值

2 个答案: