[Study Notes Dish Dog Learning C] Classic Written Exam Questions of Dynamic Memory Management

题目一：

1. 问题
   Please ask the following code,运行Test 函数会有什么样的结果？

void GetMemory(char* p) {
    
	p = (char*)malloc(100);
}
void Test(void) {
    
	char* str = NULL;
	GetMemory(str);
	strcpy(str, "hello world");
	printf(str);
}
int main() {
    
	Test();
	return 0;
}

2. 代码分析
     str传给GetMemoryThe function is passed the value,所以GetMemory函数的形参pWhat is received is a null pointer.GetMemoryThe function performs dynamic space application internally,地址存放在p中,并不会影响Test函数中的str.所以str的值依旧是NULL,所以strcpy失败.并且当GetMemory函数运行完之后,形参p销毁,dynamically developed100bytes of memory are not freed,这样就会导致内存泄露.
3. 图解
   
4. 错误更正

char* GetMemory(char* p) {
    
	p = (char*)malloc(100);
	return p;
}
void Test(void) {
    
	char* str = NULL;
	str = GetMemory(str);
	strcpy(str, "hello world");
	printf(str);
	free(str);
	str = NULL;
}
int main() {
    
	Test();
	
	return 0;
}

---

题目二

1. 问题
   Please ask the following code,运行Test 函数会有什么样的结果？

char* GetMemory(void) {
    
	char p[] = "hello world";
	return p;
}
void Test(void) {
    
	char* str = NULL;
	str = GetMemory();
	printf(str);
}
int main() {
    
	Test();

	return 0;
}

2. 代码分析
   GetMemory函数内部创建的数组是在栈区上创建的,当函数结束时,This space is also destroyed,因此pArrays have no space,返回pThe address of the array has no real meaning.如果strUse the returned address to access memory,就是非法访问.
3. 图解
   
4. 错误更正

char* GetMemory(void) {
    
	char* p = (char*)malloc(15);
	p = "hello world";
	return p;
}
void Test(void) {
    
	char* str = NULL;
	str = GetMemory();
	printf(str);
}
int main() {
    
	Test();

	return 0;
}

---

题目三

1. 问题
   Please ask the following code,运行Test 函数会有什么样的结果？

void GetMemory(char** p, int num) {
    
	*p = (char*)malloc(num);
}
void Test(void) {
    
	char* str = NULL;
	GetMemory(&str, 100);
	strcpy(str, "hello");
	printf(str);
}
int main() {
    
	Test();

	return 0;
}

2. 代码分析
   GetMemory函数获取str的地址,Then dynamically open up a space,Hand over the address of this spacestr,最后拷贝“hello”并打印,没有一点问题,But he did not perform memory release,会导致内存泄漏.

3. 图解
   

4. 错误更正

void GetMemory(char** p, int num) {
    
	*p = (char*)malloc(num);
}
void Test(void) {
    
	char* str = NULL;
	GetMemory(&str, 100);
	strcpy(str, "hello");
	printf(str);
	free(str);
	str = NULL;
}
int main() {
    
	Test();

	return 0;
}

---

题目四

1. 问题
   Please ask the following code,运行Test 函数会有什么样的结果？

void Test(void) {
    
	char* str = (char*)malloc(100);
	strcpy(str, "hello");
	free(str);
	if (str != NULL)
	{
    
		strcpy(str, "world");
		printf(str);
	}
}
int main() {
    
	Test();

	return 0;
}

2. 代码分析
   The problem with this code is that it doesn't empty,will result in a wild pointer,野指针是非常危险的.当free时,Dynamically opened space has been freed up,只不过strRemember the address of this space,Assigning a value to this space is illegal access.

3. 图解
   

4. 错误更正

void Test(void) {
    
	char* str = (char*)malloc(100);
	strcpy(str, "hello");
	free(str);
	str = NULL;
	if (str != NULL)
	{
    
		strcpy(str, "world");
		printf(str);
	}
}
int main() {
    
	Test();

	return 0;
}