I have written a piece of code that takes several integers (as many as 100 000 int) as input from a file and stores them in a "recursive" struct. As long as I run this code on my PC everything is fine. Here is the code: typedef struct node{ int data; struct node* next; } node; ... node* create(void){ node* list = (node*)malloc(sizeof(node)); return list; } node*
我写了一段代码，它将几个整数（多达100000个int）作为文件的输入，并将它们存储在“递归”结构中。 只要我在我的电脑上运行这段代码，一切都很好。 代码如下： typedef struct node{ int data; struct node* next; } node; ... node* create(void){ node* list = (node*)malloc(sizeof(node)); return list; } node* insert(node* list, int temp){ if(list == NULL){ list = create();
Is there any malloc/realloc/free like implementation where i can specify a memory region where to manage the memory allocation? I mean regular malloc (etc.) functions manages only the heap memory region. What if I need to allocate some space in a shared memory segment or in a memory mapped file? Not 100 %, As per your question you want to maintain your own memory region. so you need to go f
是否有任何malloc / realloc /免费类似的实现，我可以指定一个内存区域在哪里管理内存分配？ 我的意思是常规的malloc（等）函数只管理堆内存区域。 如果我需要在共享内存段或内存映射文件中分配一些空间，该怎么办？ 不是100％，根据你的问题你想维护自己的内存区域。 所以你需要去为自己的my_malloc ， my_realloc和my_free 实现你自己的my_malloc可能会对你有所帮助 void* my_malloc(int size) { char* ptr =
I know that the result of int *x = malloc(sizeof(int)*100); and int x[100]; is same,but the first one is allocating heap memory,and second one is allocating stack memory. Now i need to create a huge array(about 10000 element,not in a pattern),I think malloc() is more suitable. But when i ready to initialize the array, I face a problem. I cannot use any looping to init the array,how can I
我知道结果 int *x = malloc(sizeof(int)*100); 和 int x[100]; 是一样的，但第一个是分配堆内存，第二个是分配堆栈内存。 现在我需要创建一个巨大的数组（大约10000个元素，而不是模式），我认为malloc（）更合适。 但是当我准备初始化阵列时，我面临一个问题。 我不能使用任何循环来初始化数组，我如何初始化使用malloc创建的数组，就像使用一样 int x[100] = {1,2,3,4,......,6,7,5}; 当你说int a[] = { 1, 2, 3 }
I have an image compression application that now has two different versions of memory allocation systems. In the original one, malloc is used everywhere, and in the second one, I implemented a simple pool-allocator, that just allocates chunk of memory and returns parts of that memory to myalloc() calls. We've been noticing a huge memory overhead when malloc is used: At the height of its me
我有一个图像压缩应用程序，现在有两个不同版本的内存分配系统。 在原始版本中，malloc在任何地方都可以使用，而在第二个版本中，我实现了一个简单的pool-allocator，它只是分配大块内存并将该内存的一部分返回给myalloc（）调用。 当使用malloc时，我们注意到一个巨大的内存开销：在内存使用的高度，malloc（）代码需要大约170兆字节的内存用于1920x1080x16bpp的映像，而池分配器只分配48兆字节，其中47被程序使用。 就内
This is related to following question, How to Declare a 32-bit Integer in C Several people mentioned int is always 32-bit on most platforms. I am curious if this is true. Do you know any modern platforms with int of a different size? Ignore dinosaur platforms with 8-bit or 16-bit architectures. NOTE: I already know how to declare a 32-bit integer from the other question. This one is mo
这与以下问题有关， 如何在C中声明一个32位整数 几个人提到int在大多数平台上总是32位。 我很好奇，如果这是真的。 你是否知道任何具有不同大小int的现代平台？ 忽略具有8位或16位体系结构的恐龙平台。 注：我已经知道如何声明另一个问题的32位整数。 这更像是一次调查，以找出哪些平台（CPU / OS /编译器）支持其他大小的整数。 正如一些人所说，如果你想使用特定大小的变量，特别是在编写涉及位操作的代码时，你
I have a struct defined like this typedef struct { char* Value; unsigned int Length; } MY_STRUCT; I'm creating an array of these structs using calloc: MY_STRUCT* arr = (MY_STRUCT*)calloc(50, sizeof(MY_STRUCT)); Then, in a loop, I'm accessing each struct and trying to allocate and assign a value to the Value field using calloc and memcpy: int i; for(i = 0; i < 50; i++) {
我有一个像这样定义的结构 typedef struct { char* Value; unsigned int Length; } MY_STRUCT; 我使用calloc创建了这些结构的数组： MY_STRUCT* arr = (MY_STRUCT*)calloc(50, sizeof(MY_STRUCT)); 然后，在一个循环中，我访问每个结构并尝试使用calloc和memcpy分配值并将值赋给Value字段： int i; for(i = 0; i < 50; i++) { MY_STRUCT myStruct = arr[i]; int valueLength = get_value_length(i);//for s
While playing with the implementation of a hashmap toy example (for fun) I've found a strange behaviour, calloc does not initialize the entire memory block I want to zero, as supposed to do. The following code should produce no output if the entire memory block is zeroed: #include <stdio.h> #include <stdlib.h> #include <stdint.h> #include <string.h> #define DICT_INI
在玩哈希贴图玩具的例子（为了好玩）时，我发现了一个奇怪的行为，calloc并没有初始化我想要的零的整个内存块，就像应该这样做。 如果整个内存块归零，以下代码不应该输出： #include <stdio.h> #include <stdlib.h> #include <stdint.h> #include <string.h> #define DICT_INITIAL_CAPACITY 50 typedef struct dictionary_item { char* ptr_key; void* ptr_value; } dict_item; typedef s
I would like to ask you a question. I have the following code: #include <stdio.h> #include <stdlib.h> #include <unistd.h> #define XXX 1024*1024 int main() { int *p; unsigned long x=0; while (1) { //p = (int *) calloc (1,XXX); p = (int *) malloc (XXX); memset (p,0,XXX); x++;
我想问你一个问题。 我有以下代码： #include <stdio.h> #include <stdlib.h> #include <unistd.h> #define XXX 1024*1024 int main() { int *p; unsigned long x=0; while (1) { //p = (int *) calloc (1,XXX); p = (int *) malloc (XXX); memset (p,0,XXX); x++; printf ("%lu MB all
I was taught that if you do malloc(), but you don't free(), the memory will stay taken until a restart happens. Well, I of course tested it. A very simple code: #include <stdlib.h> int main(void) { while (1) malloc(1000); } And I watched over it in Task Manager (Windows 8.1). Well, the program took up 2037.4 MB really quickly and just stayed like that. I understand it's
我被告知，如果你做malloc（），但你没有释放（），那么内存将保持被占用，直到重新启动。 那么，我当然会测试它。 一个非常简单的代码： #include <stdlib.h> int main(void) { while (1) malloc(1000); } 我在任务管理器（Windows 8.1）中观看了它。 那么，该计划真的很快就占用了2037.4 MB，并且保持这种状态。 我知道这可能是Windows限制程序。 但是这里有个奇怪的部分：当我关闭控制台时，尽管我被教导
This question already has an answer here: What happens if I try to access memory beyond a malloc()'d region? 5 answers What you are doing is undefined behavior. int *p = (int *) malloc(sizeof(int)); This allocates sizeof(int) bytes starting from the address returned by malloc . When you do *(p+1) = 41; you are dereferencing to a memory location which has not been allocated on the h
这个问题在这里已经有了答案： 如果我尝试访问malloc（）'d区域之外的内存，会发生什么？ 5个答案 你在做什么是未定义的行为。 int *p = (int *) malloc(sizeof(int)); 这从malloc返回的地址开始分配sizeof(int)字节。 当你这样做 *(p+1) = 41; 您正在取消引用未在堆中分配的内存位置。 它的地址是p + sizeof(int) ，它是一个不受管理的地址。 这产生了未定义的行为， 通过观察结果可以得出每个结论都是无关紧