Question

我试图写入我使用malloc()分配的字节。我正在努力正确地打印出比特和值。

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("%016X\n", heap[i]);
        heap[i] = "AAA";
        printf("%p\n", heap[i]);
        i++;
    }
}

这是我得到的输出

0000000000000000
0xc7
0000000000000000
0xc7

Answer 1

了解C中“string”和“c”字符之间的区别。试试这段代码：

#include <stdio.h>

int main(){

    /* Usual way */
    char *a = "A";
    char *b = "B";
    char *c = "C";

    printf("Address of a = 0x%x\n",a);
    printf("Address of b = 0x%x\n",b);
    printf("Address of c = 0x%x\n",c);

    /* Explicit way - Because you asked above question */
    printf("This is Base Address of String A = 0x%x\n","A");
    printf("This is Base Address of string B = 0x%x\n","B");
    printf("This is Base Address of string C = 0x%x\n","C");

    /* Now, let us print content - The usual way */
    printf("Pointer value a has %x\n",*a);
    printf("Pointer value b has %x\n",*b);
    printf("Pointer value c has %x\n",*c);

    /* The unusual way */
    printf("Value of String A  %x\n",*"A");
    printf("Value of String B  %x\n",*"B");
    printf("Value of String C  %x\n",*"C");

}

上面的代码将生成编译器警告，因为char *被格式化为unsigned int，但只是忽略它以理解该示例。

输出如下所示：

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，你在没有真正了解其含义的情况下进行一些操作：

while (i < 2) 
{   
    printf("%016X\n", 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("%p\n", heap[i]);    // You are printing a pointer, why?
    i++;
}

C中的char只能包含一个字符。所以这有道理：

 char a = 'b';

如果你想要一个字符串，你需要一个char数组：

 char * a = "AAA";

更多阅读here

所以我会用这种方式重写代码：

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

打印和写入字节

2 个答案: