C-打印和写作字节

Question

我正在尝试写入我使用 malloc()分配的字节。我真的很努力地正确打印出零件和值。

int main(){
    unsigned char *heap = (unsigned char *) malloc( 2 * sizeof(char)); //allocate two bytes
    int n= 2, i =0;
    unsigned char* byte_array = heap;
    while (i < 2) //trying to write over the first byte then print out to verify
    {   
        printf("%016Xn", heap[i]);
        heap[i] = "AAA";
        printf("%pn", heap[i]);
        i++;
    }
}   

这是我得到的输出

0000000000000000
0xc7
0000000000000000
0xc7

Answer 1

了解C中的"字符串"和'c'字符之间的差异。尝试此代码：

#include <stdio.h>
int main(){
    /* Usual way */
    char *a = "A";
    char *b = "B";
    char *c = "C";
    printf("Address of a = 0x%xn",a);
    printf("Address of b = 0x%xn",b);
    printf("Address of c = 0x%xn",c);
    /* Explicit way - Because you asked above question */
    printf("This is Base Address of String A = 0x%xn","A");
    printf("This is Base Address of string B = 0x%xn","B");
    printf("This is Base Address of string C = 0x%xn","C");
    /* Now, let us print content - The usual way */
    printf("Pointer value a has %xn",*a);
    printf("Pointer value b has %xn",*b);
    printf("Pointer value c has %xn",*c);
    /* The unusual way */
    printf("Value of String A  %xn",*"A");
    printf("Value of String B  %xn",*"B");
    printf("Value of String C  %xn",*"C");
}

上面的代码将生成编译器警告，因为char *格式为未签名的int，但只是忽略它以理解示例。

输出看起来如下：

Address of a = 0xedfce4a
Address of b = 0xedfce4c
Address of c = 0xedfce4e
This is Base Address of String A = 0xedfce4a
This is Base Address of string B = 0xedfce4c
This is Base Address of string C = 0xedfce4e
Pointer value a has 41
Pointer value b has 42
Pointer value c has 43
Value of String A  41
Value of String B  42
Value of String C  43

Answer 2

首先，您正在进行一些操作而不真正知道含义：

while (i < 2) 
{   
    printf("%016Xn", heap[i]); // You're printing the value of heap[i] in hexadecimal that
                                // is not even setted
    heap[i] = "AAA";            // This operation has no sense, 'cause a 
                                // "char" can only contain 1 character

    printf("%pn", heap[i]);    // You are printing a pointer, why?
    i++;
}

C中的字符只能包含一个字符。因此，这有一种理解：

 char a = 'b';

，如果您想拥有一个字符串，则需要一个char的数组：

 char * a = "AAA";

以获取更多阅读此处的阅读

因此，我将以这种方式重写代码：

 while (i < 2){
     printf("First: %cn",heap[i]);
     heap[i] = 'a';
     printf("After: %cn",heap[i]);
     i++;
 }