[英]Strange error in C with a malloc of pointer in a pointer
这里我的示例代码工作正常:
//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。
现在这里我的例子有奇怪的错误:
// 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
似乎 malloc 函数在第二个指针中写入/粉碎。 我不明白我的代码哪里出了问题,很奇怪。
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),您可以安全地执行以下操作:
structDeep2->structDeep1->arrChar1 = malloc( sizeof(char ) * 2);
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