尝试显示指针值和指针地址时代码未运行

Question

我正在学习 C 语言课程，刚刚开始学习指针。我刚去完成本节的实验室，但我的代码无法正常运行。于是我打开了实验室的正确答案代码，然后我一行一行的把实验室的答案代码和我自己的答案代码进行了比较。我的一些标签是不同的，但不会导致任何错误。

我不知道为什么我的代码不能运行，但实验室的代码可以。

这是我尝试运行代码时在 xcode 中发生的情况的图片：

非常感谢任何帮助。

这是我的代码不起作用：

#include <stdio.h>

int main()
{


    int age = 40;
    float gpa = 3.25;
    char grade = 'A';
    double x = 0.000009;
    char companyName[20];

    printf("The address of age is: %d\n", &age);
    printf("The size of age is: %lu\n", sizeof(age));

    printf("The address of gpa is: %d\n", &gpa);
    printf("The size of gpa is: %lu\n", sizeof(gpa));

    printf("The address of grade is: %d\n", &grade);
    printf("The size of grade is: %lu\n", sizeof(grade));

    printf("The address of x is: %d\n", &x);
    printf("The size of x is: %lu\n", sizeof(x));

    printf("The address of companyName is: %d\n", &companyName);
    printf("The size of companyName is: %lu\n", sizeof(companyName));


    int *pointerIntAge;

    pointerIntAge = &age;

    float *pointerFloatGpa;

    pointerFloatGpa = &gpa;

    char *pointerCharGrade;
    pointerCharGrade = &grade;

    double *pointerDoubleX;

    pointerDoubleX = &x;

    char *pointerCharCompanyName;

    pointerCharCompanyName = &companyName;


    printf("The value of pointerIntAge is: %d\n", *pointerIntAge);
    printf("The value of pointerFloatGpa is: %f\n", *pointerFloatGpa);
    printf("The value of pointerCharGrade is: %c\n", *pointerCharGrade);
    printf("The value of pointerDoubleX is: %f\n", *pointerDoubleX);
    printf("The valie of pointerCharCompanyName is: %s\n", *pointerCharCompanyName); 


    printf("The address of pointerIntAge is: %p\n", pointerIntAge);
    printf("The address of pointerFloatGpa is: %p\n", pointerFloatGpa);
    printf("The address of pointerCharGrade is: %p\n", pointerCharGrade);
    printf("The address of pointer DoubleX is: %p\n", pointerDoubleX);
    printf("The address of pointerCharCompanyName is: %p\n", pointerCharCompanyName);

    *pointerIntAge+=5;
    printf("Just added 5 to pointer age\n");
    printf("The new value of age is: %d\n", age);

    printf("The value of age through pointer is: %d\n", *pointerIntAge);

    return 0;
}

这是实验室的有效代码：

#include <stdio.h>

int main ()
{
    int age = 40;
    float gpa = 3.25;
    char grade = 'A';
    double x = 0.000009;
    char companyName[20];

    printf("Address of age: %d\n", &age);
    printf("Size of age: %lu\n", sizeof(age));
    printf("Address of GPA: %d\n", &gpa);
    printf("Size of age: %lu\n", sizeof(gpa));
    printf("Address of grade: %d\n", &grade);
    printf("Size of grade: %lu\n", sizeof(grade));
    printf("Address of x: %d\n", &x);
    printf("Size of x: %lu\n", sizeof(x));
    printf("Address of companyName: %d\n", &companyName);
    printf("Size of companyName: %lu\n", sizeof(companyName));

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

    printf("\nValue of Age though pointer: %d", *pAge);
    printf("\nValue of GPA though pointer: %0.2f", *pGpa);
    printf("\nValue of Grade though pointer: %c", *pGrade);
    printf("\nValue of X though pointer: %f", *pX);
    printf("\nValue of Company Name though pointer: %s\n", *pCompanyName);

    printf("\nThe address from Age pointer: %p", pAge);
    printf("\nThe address from Gpa pointer: %p", pGpa);
    printf("\nThe address from Grade pointer: %p", pGrade);
    printf("\nThe address from X pointer: %p", pX);
    printf("\nThe address from companyName pointer: %p\n", pCompanyName);

    *pAge += 5; //Added 5 to Age through pointer.
    printf("\nValue of Age: %d", age);
    printf("\nValue of Age through pointer: %d", *pAge);

    return 0;
}

Answer 1

companyName 未初始化，因此printf("The valie of pointerCharCompanyName is: %s\n", *pointerCharCompanyName); 将打印垃圾（如果其中某处有 NULL）或超出范围（如果没有）。

这是因为 C 中的字符串（实际上只是字符数组）应该以空值结尾。这意味着它们应该在末尾有一个 '\0' 字符 - 对字符串进行操作的方法使用它来了解字符串的结束位置。

试试char companyName[20] = { 0 }; 或char companyName[20] = { "some string" };

不过，还有一些其他问题需要解决。

1. 使用%p 打印地址，而不是%d。
2. 如果您有编译器警告，您会注意到 pointerCharCompanyName = &companyName; 和 printf("The valie of pointerCharCompanyName is: %s\n", *pointerCharCompanyName); 都存在由于对指针理解不足而导致的问题。
3. 如果应该工作的示例代码真的是示例代码，它也会出现这些问题，您应该找到一个新类。

Answer 2

你的问题是这一行：pCompanyName = &companyName;

您需要打开编译器警告。你应该得到一个警告“来自不兼容的指针类型的赋值”。

companyName 已经是一个指针。因此，您的行尝试将指针分配给指针的地址（即指向指针的指针）。

应该是：pCompanyName = &companyName[0]; 或简单的pCompanyName = companyName;

另外，您在 printf("The address of age is: %d\n", &age); 等行中使用 %d 而不是 %p

另外两个版本的代码都错误地使用 %lu 而不是 %u 来表示 sizeof（它返回的 size_t 是 unsigned int，而不是 unsigned long）。

实验室代码也是错误的。根据内存的分配方式，它是否崩溃是愚蠢的运气。您还存在使用 %s 打印没有值的字符串的问题，因此随机可能有也可能没有空终止符。您还需要将带有空终止符的字符放入公司名称中。类似companyName[20] = "test";