Question

我在从文件加载数据时遇到问题。这段代码用一些随机数据填充高分（仅用于调试）

int i = 0; 
for (i = 0; i < MAX_HIGH_SCORE; i++)
{
    high_score_name[i] = al_ustr_new("123456789012345678901");
    high_score[i] = 1000*i;
}

之后我将所有这些保存到文件中。

ALLEGRO_FILE* high_score_file = al_fopen("hs.txt","wb");
for (i = 0; i < MAX_HIGH_SCORE; i++)
{
    int buffer = al_ustr_size(high_score_name[i]);
    al_fwrite32le(high_score_file, buffer);
    al_fwrite(high_score_file, al_cstr(high_score_name[i]), buffer);
    al_fwrite(high_score_file, &high_score[i], sizeof(high_score[i]));

}
al_fclose(high_score_file);

它运作得很好。它有四个字节为大小保留，然后是high_score_name[]的21个字符，后跟high_score[] int，也占用4个字节。

当我尝试加载时出现问题。我试图在很多方面做，但它不会编译，因为行

char* buffer= al_malloc(size); 给出错误：

类型“void *”的值不能用于初始化“char *
类型的实体

for (i = 0; i < MAX_HIGH_SCORE; i++)
{
    int size = al_fread32le(high_score_file);
    char* buffer = al_malloc(size);
    al_fread(high_score_file, buffer, size);
    high_score_name[i] = al_ustr_new_from_buffer(buffer, size);

    int *buffer2 = &high_score[i];          
    al_fread(high_score_file, buffer2, sizeof(high_score[i]));

    al_free(buffer);
}

al_fclose(high_score_file);

我坚持了两天，我真的会帮助你。

Answer 1

malloc的输出类型为void *，你应该尝试显式地转换为char *。

char* buffer = static_cast<char *>(al_malloc(size))

Answer 2

如果你的函数在内部使用malloc来分配内存，那么malloc返回一个void *，如果你的al_malloc返回void * ...使用类型转换为conert将它转换为char *。

char * buffer =（char *）al_malloc（size）;

这应该有用。

malloc的函数原型：

  void *malloc(size_t size);

创建缓冲区以加载数据时出错

2 个答案: