C 使用指针从方法中获取字符数组值。

我一直在努力实现以下目标： 我的目标是使用main（）中的指针访问在方法（）中创建的元素

我想打印缓冲区值的每个元素，就像方法（）使用ptr创建的一样。有没有关于我如何做这件事的建议

我不确定我是否误解了什么，但我认为ptr指向缓冲区的地址。因此，取消引用会给我缓冲区[0]

---

谢谢。

这是您的代码的固定和注释版本。在评论中询问是否有smth。你不明白

#include <stdio.h>
#include <stdlib.h>

// takes in address of pointer

//Hex: 0xab is larger than the max value of a signed char.
//Most comilers default to signed char if you don't specify unsigned.
//So you need to use unsigned for the values you chose

int method(unsigned char** input) { //<<< changed
    unsigned char *buffer = malloc(sizeof(char)*10);

    //Check for malloc success  <<< added
    if(!buffer)
        exit(EXIT_FAILURE);

    buffer[0] = 0x12;
    buffer[1] = 0x34;
    buffer[2] = 0xab;
//I recommend not to mix array notation and pointer notation on the same object.
//Alternatively, you could write:

*buffer = 0x12;
*(buffer + 1) = 0x34;
*(buffer + 2) = 0xab;

    //buffer already contains the address of your "array".
    //You don't want the address of that address
    *input = buffer;  //<<< changed (removed &)

    printf("%x\n", *buffer); // this prints 0x12
    //Not casting &buffer will likely work (with compiler warnings
    //But it is better to conform. Either use (char *) or (void *)
     //<<< added the cast for printf()
    printf("%p\n", (char *)&buffer); // this prints address of buffer example: 0x7fffbd98bf78
    printf("%p\n", *input); // this prints address of buffer

    return 0;
}

int main(){
    unsigned char *ptr;
    method(&ptr);

    printf("%p\n", ptr); // this prints address of buffer

    //this does not seem to print out buffer[0]
    for(int i = 0; i < 3; i++){
       //<<< changed to obtain content of buffer via ptr for loop.
        unsigned char buf_elem = *(ptr + i);
        printf("buffer[%d] in hex: %x\t in decimal: %d\n", i, buf_elem, buf_elem);
}

// Don't forget to free the memory.  //<<< changed
free(ptr);
}

#包括
#包括
//接收指针的地址
//十六进制：0xab大于有符号字符的最大值。
//如果不指定unsigned，大多数Comiler默认为signed char。
//因此，您需要对所选的值使用unsigned
int方法（unsigned char**input）{/
&buffer
是指向变量的指针，而不是它所指向的位置（这将是普通的
buffer
）。缓冲区
的类型是
char**
，而
*input
的类型是
char*
。此外，如果要使用
printf
打印指针，您首先需要将其强制转换为
void*
，然后使用
%p”
格式。如果您在答案中添加更改的内容以及更改的原因，将更有帮助。
#include <stdio.h>
#include <stdlib.h>

// takes in address of pointer

//Hex: 0xab is larger than the max value of a signed char.
//Most comilers default to signed char if you don't specify unsigned.
//So you need to use unsigned for the values you chose

int method(unsigned char** input) { //<<< changed
    unsigned char *buffer = malloc(sizeof(char)*10);

    //Check for malloc success  <<< added
    if(!buffer)
        exit(EXIT_FAILURE);

    buffer[0] = 0x12;
    buffer[1] = 0x34;
    buffer[2] = 0xab;
//I recommend not to mix array notation and pointer notation on the same object.
//Alternatively, you could write:

*buffer = 0x12;
*(buffer + 1) = 0x34;
*(buffer + 2) = 0xab;

    //buffer already contains the address of your "array".
    //You don't want the address of that address
    *input = buffer;  //<<< changed (removed &)

    printf("%x\n", *buffer); // this prints 0x12
    //Not casting &buffer will likely work (with compiler warnings
    //But it is better to conform. Either use (char *) or (void *)
     //<<< added the cast for printf()
    printf("%p\n", (char *)&buffer); // this prints address of buffer example: 0x7fffbd98bf78
    printf("%p\n", *input); // this prints address of buffer

    return 0;
}

int main(){
    unsigned char *ptr;
    method(&ptr);

    printf("%p\n", ptr); // this prints address of buffer

    //this does not seem to print out buffer[0]
    for(int i = 0; i < 3; i++){
       //<<< changed to obtain content of buffer via ptr for loop.
        unsigned char buf_elem = *(ptr + i);
        printf("buffer[%d] in hex: %x\t in decimal: %d\n", i, buf_elem, buf_elem);
}

// Don't forget to free the memory.  //<<< changed
free(ptr);
}