Question

这里我的示例代码很好用：

  //define struct
struct myStruct {
   char     *arrChar1;
   char     *arrChar2;
}
// code in Main
struct myStruct *structDeep1 = malloc( sizeof( struct myStruct *) );
structDeep1->arrChar1 = NULL;
structDeep1->arrChar2 = NULL;
structDeep1->arrChar1 = malloc( sizeof(char ) * 2); //2 char

if( structDeep1->arrChar1 == NULL)  puts("trace error");
else                                    puts("trace ok")
//stdout -> trace OK

没问题。

现在我的例子有一个奇怪的错误：

// define a second struct
struct myStructDeep2{
    struct myStruct     *structDeep1;
}
// next code in Main
struct myStructDeep2 structDeep2 = malloc( sizeof( struct myStructDeep2*) );
structDeep2->structDeep1 = malloc( sizeof( struct myStruct *) );
structDeep2->structDeep1->arrChar1 = NULL;
structDeep2->structDeep1->arrChar2 = NULL;
structDeep2->structDeep1->arrChar1 = malloc( sizeof(char ) * 2); //2 char

if( structDeep2->structDeep1->arrChar1 == NULL)     puts("trace error");
else                                            puts("trace ok")
//stdout -> trace error

在第二个指针中似乎是malloc函数write / crush。我不明白我的代码中的问题在哪里，这很奇怪。

Answer 1

当你这样做时：

structDeep2->structDeep1 = malloc( sizeof( struct myStruct *) );

你分配了一个指针的大小，但你想分配一个结构，所以你必须这样做：

structDeep2->structDeep1 = malloc( sizeof( struct myStruct) );

现在（假设malloc成功），你可以安全地做到：

structDeep2->structDeep1->arrChar1 = malloc( sizeof(char ) * 2);

C中带有指针malloc的奇怪错误

1 个答案: