I have set stack size to 2000Kb by ulimit -s 2000 and ulimit -Ss 2000 for hard limit. And in the below program i have allocated appox 2040000(510000 x 4) bytes which is less than i limited ie,. 2048000(2000*4)bytes but i see that my program crashes! Can anybody suggest why this happens. #include <stdio.h> #include <malloc.h> int main() { int a[510000] = {0}; a[510000] = 1;
我已经通过ulimit -s 2000设置堆栈大小为2000Kb，ulimit -Ss 2000为硬限制。 而在下面的程序中，我已经分配了小于我限制的appox 2040000（510000×4）字节，即，。 2048000（2000 * 4）字节，但我看到我的程序崩溃！ 有人可以提出为什么会发生这种情况 #include <stdio.h> #include <malloc.h> int main() { int a[510000] = {0}; a[510000] = 1; printf("%d", a[510000]); fflush(stdout); s
For instance, lets suppose that instead of buffers growing in the opposite direction of the stack, they grow in the same direction. If I have a character buffer containing the string "Hello world", instead of 'H' being placed at the lowest address, it is placed at the highest address, and so on. If an input string copied to a buffer were to overflow it could not overwrite the
例如，让我们假设，而不是缓冲区在堆栈的相反方向上生长，而是沿着相同的方向增长。 如果我有一个包含字符串“Hello world”的字符缓冲区，而不是将“H”放置在最低地址处，则将其放置在最高地址处，依此类推。 如果复制到缓冲区的输入字符串溢出，它不能覆盖该函数的返回地址，但肯定还有其他东西可以覆盖。 我的问题是 - 如果输入的字符串足够长，什么东西可以被覆盖？ 堆和堆栈之间是否存在可被覆盖的库函数？ 堆变量可以被
Let's assume that I have a struct with a void ** member. This member serves as an array of pointers to data channels. The data type is irrelevant. Below is an example of how I'd like to allocate memory for this 2D array and then associate channel pointers with their memory. #define N_CHANNELS (/*Number of data channels*/) #define N_SAMPLES (/*Number of data samples per channel*/) type
假设我有一个void **成员的结构。 这个成员可以作为数据通道的指针数组。 数据类型是不相关的。 以下是我想如何为这个二维数组分配内存然后将通道指针与其内存相关联的示例。 #define N_CHANNELS (/*Number of data channels*/) #define N_SAMPLES (/*Number of data samples per channel*/) typedef /*Some data type*/ DATA_TYPE; void **ppdata; /*Conceptual struct member*/ /*Allocate memory for array of data chan
I am having a difficult time fully understanding what is going on. This is how I am reading the code at the bottom. Define A as a pointer to an address of a pointer that points to address of a double. Allocate 4 blocks of memory on the heap that can each hold an address to a double. Return the address of the first block allocated to A. Allocate 6 blocks of memory on the heap that can eac
我很难完全理解正在发生的事情。 这是我阅读底部代码的方式。 将A定义为指向指向double地址的指针地址的指针。 在堆上分配4块内存，每块内存可以将地址保存为一倍。 返回分配给A的第一个块的地址 在堆上分配6个内存块，每个内存块可以保存一个双数据类型，并为之前分配的4个块中的每个块返回第一个块的地址。 变量A仍然包含四个指针中第一个块的地址，前提是A [0]改变了A指向的地址的值而不是A本身。 现在，这是图
For the C statement given below i would like to know where the memmory allocation will take place. char* ptr="Hello";//ptr is a automatic variable then the pointer variable ptr will be allocated on the stack,but where will this string "Hello" be allocated. Is it on Stack or on Heap? And what about memory allocation for initialization statement like char ptr[]="Hello"; Th
对于下面给出的C语句，我想知道内存分配将在哪里发生。 char* ptr="Hello";//ptr is a automatic variable 那么指针变量ptr将被分配到栈上，但是这个字符串“Hello”将被分配到哪里。 它是堆栈还是堆？ 那么对于像char ptr [] =“Hello”这样的初始化语句的内存分配呢？ 标准没有说（它不知道“堆栈”，“堆”等）。 但实际上答案是：都不是。 字符串文字将被存储在数据部分中，通常是只读页面。 作为一个侧面说明，正如Als在
In C++ it's actually possible to throw an exception by value without allocating memory on a heap, so this situation makes sense. But in .NET framework OutOfMemoryException is a reference type, therefore it is allocated on a heap. How does .NET framework allocates memory for OutOfMemoryException when there is not enough memory to create a new object? It is preallocated by the runtime. If
在C ++中，实际上可以通过值来引发异常，而不必在堆上分配内存，所以这种情况是有道理的。 但是在.NET框架中， OutOfMemoryException是一个引用类型，因此它被分配在一个堆上。 如果没有足够的内存来创建新对象，.NET框架如何为OutOfMemoryException分配内存？ 它由运行时预先分配。 如果您探索任何托管过程的堆，您会发现该异常的一个实例。 以下是Hello World应用程序的预分配例外情况： 0:003> !dumpheap -stat -ty
As Value type variable allocates memory in Stack where as Reference Type allocates it in Heap. So how the memory allocated when a value type variable(eg int i =4;) is declared in the reference type (eg. in a class). How the overall memory allocation works in .net for value type & reference typ, and also value type inside the scope of refence type. Please explain it or provide any links
由于值类型变量在堆栈中分配内存，其中参考类型将其分配到堆中。 那么如何在引用类型（例如，类中）中声明值类型变量（例如int i = 4;）时分配的内存。 整个内存分配如何在.net中用于值类型和引用类型，以及值类型内的值类型。 请解释它或提供有关该链接的任何链接。 谢谢 那么如何在引用类型（例如，类中）中声明值类型变量（例如int i = 4;）时分配的内存。 如果对象位于堆上，则表示它的所有成员变量都位于此处。
I am trying to answer a quiz regarding the following code: struct B { int _arr[5]; }; struct A{ struct B* _b; }; #include <stdlib.h> int main(){ struct A arr[5]; struct A * a = (struct A*)malloc(sizeof (struct A)); return 0; } I am trying to understand how much memory (in bytes) is allocated on the program's stack. So I first used the following (and probably wrong) log
我正在尝试回答关于以下代码的测验： struct B { int _arr[5]; }; struct A{ struct B* _b; }; #include <stdlib.h> int main(){ struct A arr[5]; struct A * a = (struct A*)malloc(sizeof (struct A)); return 0; } 我想了解程序堆栈上分配了多少内存（以字节为单位）。 所以我首先使用了下面的（也许是错误的）逻辑： 计算结构B的大小= sizeof（int）* 5 = 20; 计算结构A的大小= sizeof（B）= 2
This simple method just creates an array of dynamic size n and initializes it with values 0 ... n-1. It contains a mistake, malloc() allocates just n instead of sizeof(int) * n bytes: int *make_array(size_t n) { int *result = malloc(n); for (int i = 0; i < n; ++i) { //printf("%d", i); result[i] = i; } return result; } int main() { int *result = make_arr
这个简单的方法只是创建一个动态大小为n的数组，并用值0 ... n-1初始化它。 它包含一个错误，malloc（）只分配n而不是sizeof（int）* n个字节： int *make_array(size_t n) { int *result = malloc(n); for (int i = 0; i < n; ++i) { //printf("%d", i); result[i] = i; } return result; } int main() { int *result = make_array(8); for (int i = 0; i < 8; ++i) {
Possible Duplicate: Why isn't sizeof for a struct equal to the sum of sizeof of each member? I have the next code: http://ideone.com/brmRy #include <stdio.h> #include <stdlib.h> typedef struct Test { int a; int b; char c; } Test; int main(void) { Test *obj = (Test*)malloc(sizeof(Test)); printf("Size results:rnrnstruct: %irnint #1:
可能重复： 为什么不是sizeof等于每个成员的sizeof之和？ 我有下一个代码： http://ideone.com/brmRy #include <stdio.h> #include <stdlib.h> typedef struct Test { int a; int b; char c; } Test; int main(void) { Test *obj = (Test*)malloc(sizeof(Test)); printf("Size results:rnrnstruct: %irnint #1: %irnint #2: %irnchar #1: %irn",