分割断层的strcpy() [英] Segmentation Fault in strcpy()
问题描述
我有一个基本的结构,这样
I have a basic structure like this
typedef struct struck {
char* id;
char* mat;
int value;
char* place;
} *Truck;
和机能缺失这样它创建该结构的一个新的实例:
And afunction like this which creates a new "instance" of that struct:
Truck CTruck(char* id, char* mat, int value, char* place) {
Truck nT = (Truck) malloc(sizeof (Truck));
nT->value = value;
strcpy(nT->id, id);
strcpy(nT->mat, mat);
strcpy(nT->place, place);
return nT;
}
我收到一个错误在第一个的strcpy
。它编译没有问题。
推荐答案
您的typedef定义汽车
为结构来袭*
,即一个指针。所以,它的大小将是 4
或 8
依赖于架构,而不是结构的大小
Your typedef defines Truck
as a struct struck *
, i.e. a pointer. So it's size will be 4
or 8
depending on the architecture and not the size of the struct
使用的sizeof(*卡车)
来获得结构的实际大小。
Use sizeof(*Truck)
to get the actual size of the struct.
您还需要为角色分配内存。最简单的方法将使用的strdup()
。
You also need to allocate memory for the characters. The easiest way would be using strdup()
.
Truck CTruck(const char* id, const char* mat, int value, const char* place) {
Truck nT = malloc(sizeof (*Truck));
nT->value = value;
nT->id = strdup(id);
nT->mat = strdup(mat);
nT->place = strdup(place);
return nT;
}
不过,我建议改变你的typedef所以它的结构的别名,而不是一个指向它:
However, I would suggest changing your typedef so it's an alias for the struct, not for a pointer to it:
typedef struct {
char* id;
char* mat;
int value;
char* place;
} Truck;
在你的函数,你再使用这样的:
In your function you then use this:
Truck *nT = malloc(sizeof(Truck));
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