[英]Initialization function (i.e. “constructor”) for struct, why are variables not being initialized properly?
I have an unnamed structure called `FooStruct', and I wrote a function for it that initializes all its variables and also takes care of dynamic memory allocation. 我有一个名为“ FooStruct”的未命名结构,我为此编写了一个函数,该函数初始化其所有变量,并负责动态内存分配。
I tried running this code, but it is not producing the results that I am expecting. 我尝试运行此代码,但未产生预期的结果。 For some reason, the variables are initialized correctly when inside the function, but then they change again once the block is exited.
由于某些原因,变量在函数内部时已正确初始化,但是一旦退出该块,它们便会再次更改。
A project that I'm working on requires that I use unnamed structs only. 我正在处理的项目要求仅使用未命名的结构。 Why is this happening?
为什么会这样呢?
#include <stdio.h>
#include <stdlib.h>
typedef struct {
int fooPrimitive;
} FooStruct;
void FooStructInit(FooStruct * ptr, int num) {
ptr = (FooStruct*)malloc(sizeof(FooStruct));
ptr -> fooPrimitive = num;
// Expected output: 5
// Actual output: 5
printf("ptr -> fooPrimitive: %d\n", (ptr -> fooPrimitive));
}
int main() {
FooStruct * ptr;
FooStructInit(ptr, 5);
// Expected output: 5
// Actual output: some random number
printf("FooStruct -> fooPrimitive: %d\n", (ptr -> fooPrimitive));
}
When you pass ptr
to FooStructInit
, it is copied and this copy is assigned the return value of malloc
then. 当将
ptr
传递给FooStructInit
,它将被复制,然后为该副本分配malloc
的返回值。 This is called pass-by-value . 这称为传递值 。 Instead, pass a pointer to
ptr
to actually write to ptr
and not just the passed argument: 而是传递指向
ptr
的指针以实际写入ptr
,而不仅仅是传递的参数:
void FooStructInit(FooStruct** ptr, int num) {
*ptr = malloc(sizeof(FooStruct));
(*ptr)->fooPrimitive = num;
// Expected output: 5
// Actual output: 5
printf("(*ptr)->fooPrimitive: %d\n", ((*ptr)->fooPrimitive));
}
int main() {
FooStruct* ptr;
FooStructInit(&ptr, 5);
// Expected output: 5
// Actual output: 5
printf("FooStruct->fooPrimitive: %d\n", (ptr->fooPrimitive));
}
Notes: 笔记:
free
after malloc
malloc
之后调用free
malloc
malloc
的结果 C only has pass by value semantics, so when you pass a pointer to a function, the original value of the pointer is not changed, only the local copy is changed in the function. C仅具有按值传递语义,因此,当您将指针传递给函数时,指针的原始值不会更改,而只会在函数中更改本地副本。
The way to get around this is to pass a pointer to a pointer, so that you can change the value at the original memory location. 解决此问题的方法是将一个指针传递给一个指针,以便您可以更改原始存储位置的值。
void FooStructInit(FooStruct **ptr, int num) {
*ptr = malloc(sizeof(FooStruct));
...
}
int main(void) {
FooStruct * ptr;
FooStructInit(&ptr, 5); /* get the address of the pointer */
...
}
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