1

所以我做了这个函数来接收未知数量的字符串并将它们添加到字符串数组中。

char **receiveCode(int socket){
    int line = 0;
    size_t lines = 1;
    size_t size = 1;
    char** code = malloc(sizeof(char)*size);

    while(1){
        package_struct *aPackage = receivePackage(socket);
        if(aPackage->type=='F'){break;}
        size = size + aPackage->size;
        code = realloc(code, size);
        code[line] = malloc(sizeof(char)*aPackage->size);
        strcpy(code[line],aPackage->package);
        line++;
        lines++;
        free(aPackage);
    }
    code = realloc(code, size + 2);
    code[line] = malloc(sizeof(char)*3);
    code[lines]=NULL;
    return code;
}

有时当我运行此代码时，我会收到以下错误

* 检测到 glibc./pp: realloc(): 无效的下一个大小: 0x00007f0f88001220 * *

根据 Valgrind 的说法，这发生在那个函数中。

可能我使用了太多的 malloc 和 realloc ......不过不确定。

4

3 回答 3

3

我认为问题是这样的：

char** code = malloc(sizeof(char)*size);

它应该是char *而不是char里面sizeof()

char** code = malloc(sizeof(char*)*size);

由于code是指向字符串的指针，因此为指针分配内存char*

也有同样的问题realloc

于 2012-11-21T20:15:02.527 回答
3

我假设这是分配一个数组char*：

code = realloc(code, size);

应该

code = realloc(code, size * sizeof(char*));
// and this one too
code = realloc(code, size + 2 * sizeof(char*));

另外，你不需要这个：

char** code = malloc(sizeof(char)*size);

如果你调用realloc(NULL, size)它相当于malloc(size)

size_t size = 0;
char** code = NULL;
...
code = realloc(code, size * sizeof(char*));

注意：lines对我来说似乎没用，实际上在最后两行中你覆盖了你刚刚分配的内存line==lines

于 2012-11-21T20:16:31.317 回答
1

这是一个strdup()用于简化每个新文本行的内存分配的版本。它还使用“x”版本的内存分配函数来简化内存不足错误处理（一个有点常见的习惯用法，即使是非标准的）。

所以真正剩下的所有复杂性（最终不会太多）在于管理字符串指针数组的增长。我认为这使得将处理每个字符串与处理指针数组分开变得更容易。原始代码混淆了这两个领域。

// these variants allocate memory, but abort program on failure
//  for simplified error handling - you may need different
//  error handling, but often this is enough
//
//  Also, your platform may or may not already have these functions
//      simplified versions are in the example.

void* xmalloc( size_t size);
void* xrealloc(void* ptr, size_t size);
char* xstrdup(char const* s);


char** receiveCode(int socket){
    size_t lines = 0;
    char** code = xmalloc( (lines + 1) * sizeof(*code));

    *code = NULL;

    while(1){
        package_struct *aPackage = receivePackage(socket);
        if(aPackage->type=='F') {
            free(aPackage); // not 100% sure if this should happen here or not.
                            // Is a `package_struct` with type 'F' dynamically
                            // allocated or is a pointer to a static sentinel 
                            // returned in this case?
            break;
        }


        // why use `aPackage->size` when you use `strcpy()` to
        //  copy the string anyway? Just let `strdup()` handle the details
        //
        // If the string in the `pckage_struct` isn't really null terminated, 
        // then use `xstrndup(aPackage->package, aPackage->size);` or something
        // similar.

        char* line = xstrdup(aPackage->package);
        ++lines;

        // add another pointer to the `code` array
        code = xrealloc(code, (lines + 1) * sizeof(*code));
        code[lines-1] = line;
        code[lines] = NULL;

        free(aPackage);
    }

    return code;
}


void* xmalloc(size_t size)
{
    void* tmp = malloc(size);

    if (!tmp) {
        fprintf(stderr, "%s\n", "failed to allocate memory.\n";
        exit(EXIT_FAILURE);
    }

    return tmp;
}

void* xrealloc(void *ptr, size_t size)
{
    void* tmp = realloc(ptr, size);

    if (!tmp) {
        fprintf(stderr, "%s\n", "failed to allocate memory.\n";
        exit(EXIT_FAILURE);
    }

    return tmp;
}


char* xstrdup(char const* s)
{
    char* tmp = strdup(s);

    if (!tmp) {
        fprintf(stderr, "%s\n", "failed to allocate memory.\n";
        exit(EXIT_FAILURE);
    }

    return tmp;
}

另外，我认为应该澄清它aPackage->package是字符串指针还是char[]保存字符串数据的实际位置（即，应该&aPackage->package传递给strcpy()/ xstrdup()？）。如果它真的是一个指针，它应该在之前被释放aPackage吗？

于 2012-11-21T21:21:35.247 回答