通過取消引用指針獲取值

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; 
} 

當我運行這段代碼時，Xcode回覆了大量的警告和錯誤。當聲明和初始化除age之外的所有變量的指針時，它表示incompatible pointer types。當試圖將它們打印出來時，它說除了age,format specifies a different type。爲什麼是這樣？而在代碼運行時，因爲是我得到如下結果：

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

Answer 1

您已經聲明它們作爲指向整數，而他們應該是一個指向int，一個float，一個char等上！

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

Answer 2

下面是正確的代碼

#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; 
} 

輸出是

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

你必須使用corect指針類型和格式說明在printf函數。

或者，如果你想有一個指向一個字符數組，那麼你可以寫

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

，並相應

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