C程序，为什么将char指针解引用为指针却未获得期望值 [英] C program, Why dereferece a char pointer to a pointer doesnt get expected value

问题描述

在此程序中，我具有char变量a，b是指向a的指针，而c是指向b的指针. *a=b时，*c不等于b.我不明白为什么，有人可以解释吗?

In this program I have char variable a, b is a pointer to a and c is a pointer to b. While *a=b, *c not equal to b. I don't understand why ,Can anyone explain?

我不了解的另一件事是，如果我将变量从char更改为int，则取消引用c结果b值. *c等于b，但是如果变量是char类型，则不是.

Another thing I don't understand I that if I change variable from char to int, dereference c result b value. *c equal to b.But if variable is char type, it does not.

#include<stdio.h>
#include<string.h>

int main()
{
char a =  "a" ;
char *b;


b = &a;
printf("%d\n", b);
printf("%d\n\n", &a);
printf("Deference b* hold value: %d\n", *b);
printf("a hold value: %d\n\n", a);
char *c;
c = &b;
printf("%d\n", c);
printf("%d\n\n", &b);
printf("Deference *c hold value: %d\n", *c);
printf("b hold value: %d\n\n", b);// why *c not equal b
return 0;

}

推荐答案

首先，

 char a =  "a" ;

是非法的，您实际上是在试图将指针存储到char中.您需要的是

is illegal, you're essentially trying to store a pointer into a char. What you need is

char a =  'a' ;

然后说

printf("%d\n", b);
printf("%d\n\n", &a);  //and all later pointer related prints....

在将错误类型的参数传递给%d时，会导致未定义的行为.要打印指针，您需要

causes undefined behavior as you're passing wrong type of arguments to %d. To print pointers, you need to

• 使用%p格式说明符.
• 将参数传递给void*

• use %p format specifier.
• cast the argument to void*

之后，

char *c;
c = &b;

也是错误的，请参见数据类型. &b是指向char的指针.这与已假定 不同，它与char *不同.您需要c属于char **类型.

is also wrong, see the data types. &b is a pointer to pointer-to-char. That is not the same as char *, as you have assummed. You need c to be of type char **.

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