calloc

Does the pointer passed to free() have to point to beginning of the memory block, or can it point to the interior?

假如想象 提交于 2019-11-28 04:00:15
问题 The question is in the title... I searched but couldn't find anything. Edit: I don't really see any need to explain this, but because people think that what I'm saying makes no sense (and that I'm asking the wrong questions), here's the problem: Since people seem to be very interested in the "root" cause of all the problem rather than the actual question asked (since that apparently helps things get solved better, let's see if it does), here's the problem: I'm trying to make a D runtime

Proper way to initialize C++ structs

蹲街弑〆低调 提交于 2019-11-28 02:56:54
Our code involves a POD (Plain Old Datastructure) struct (it is a basic c++ struct that has other structs and POD variables in it that needs to get initialized in the beginning.) Based one what I've read , it seems that: myStruct = (MyStruct*)calloc(1, sizeof(MyStruct)); should initialize all the values to zero, as does: myStruct = new MyStruct(); However, when the struct is initialized in the second way, Valgrind later complains "conditional jump or move depends on uninitialised value(s)" when those variables are used. Is my understanding flawed here, or is Valgrind throwing false positives?

存储空间的动态分配与释放

独自空忆成欢 提交于 2019-11-28 02:12:55
1 ,关于 malloc() 函数 函数调用的一般形式: (类型标识符 * ) malloc(size) malloc() 函数的功能是在内存动态分配一个长度为 size 的一个连续空间,含数返回值是该区域的首地址。 (类型标识符 * )是强制类型转换。因为函数返回的指针的是五类型的,用户根据存储空间的用途把函数调用返回的指针强制转换为相应的类型。 size 是一个无符号数,单位是字节。 2 , calloc() 函数 函数调用的一般格式: (类标识符 * ) calloc(n,size) calloc() 函数的功能是在内存动态分配 n 个长度为 size 的连续空间,函数返回值是该区域的首地址。 为正整数。 3 , free() 函数 函数调用一般形式: Free( 指针变量名 ); 代码 #include " iostream " using namespace std; float average( int * p, int n) { int i; float ave = 0 ; for (i = 0 ;i < n;i ++ ) ave +=* (p + i); return ave / n; 10 11 } int main() { int i,n, * p; cin >> n; p = ( int * )malloc(n * sizeof ( int )); // 或p

is it necessary to type-cast malloc and calloc [duplicate]

强颜欢笑 提交于 2019-11-28 00:59:49
问题 Possible Duplicate: Do I cast the result of malloc? I was googling to find out the reason for type-casting of malloc and calloc . But, i only found type-casting of malloc is not necessary since it return void pointer but, what about calloc . This is the same reason for calloc too ??? Now, if we move back to first point, about return value of malloc and calloc . Then, i found that, both are returning the allocated spaces . So, i'm little bit confused here. So, my questions are What is the

Why calloc takes two arguments while malloc only one?

a 夏天 提交于 2019-11-27 23:57:05
IMO one is enough, why does calloc require to split it into two arguments? I'd guess that this is probably history and predates the times where C had prototypes for functions. At these times without a prototype the arguments basically had to be int , the typedef size_t probably wasn't even yet invented. But then INTMAX is the largest chunk you could allocate with malloc and splitting it up in two just gives you more flexibility and allows you to allocate really large arrays. Even at that times there were methods to obtain large pages from the system that where zeroed out by default, so

Invalid application of sizeof to incomplete type with a struct

核能气质少年 提交于 2019-11-27 22:43:22
I have a struct where I put all the information about the players. That's my struct: struct player{ int startingCapital; int currentCapital; int startingPosition; int currentPosition; int activePlayer; int canPlay; }; And that's my main: #include <stdio.h> #include <stdlib.h> #include "header.h" int main(int argc, char *argv[]) { int s,i,numOfPlayers; struct player *players; printf("Give the number of players: \n"); scanf("%d",&numOfPlayers); players = (struct player *)calloc(numOfPlayers,sizeof(struct player)); system("PAUSE"); return 0; } I'm asking the user to give the number of players and

Proper usage of realloc()

落爺英雄遲暮 提交于 2019-11-27 04:59:58
From man realloc:The realloc() function returns a pointer to the newly allocated memory, which is suitably aligned for any kind of variable and may be different from ptr, or NULL if the request fails. So in this code snippet: ptr = (int *) malloc(sizeof(int)); ptr1 = (int *) realloc(ptr, count * sizeof(int)); if(ptr1 == NULL){ //reallocated pointer ptr1 printf("Exiting!!\n"); free(ptr); exit(0); }else{ free(ptr); //to deallocate the previous memory block pointed by ptr so as not to leave orphaned blocks of memory when ptr=ptr1 executes and ptr moves on to another block ptr = ptr1; /

Why calloc takes two arguments while malloc only one?

时光毁灭记忆、已成空白 提交于 2019-11-27 04:42:41
问题 IMO one is enough, why does calloc require to split it into two arguments? 回答1: I'd guess that this is probably history and predates the times where C had prototypes for functions. At these times without a prototype the arguments basically had to be int , the typedef size_t probably wasn't even yet invented. But then INTMAX is the largest chunk you could allocate with malloc and splitting it up in two just gives you more flexibility and allows you to allocate really large arrays. Even at that

Invalid application of sizeof to incomplete type with a struct

∥☆過路亽.° 提交于 2019-11-27 04:35:57
问题 I have a struct where I put all the information about the players. That's my struct: struct player{ int startingCapital; int currentCapital; int startingPosition; int currentPosition; int activePlayer; int canPlay; }; And that's my main: #include <stdio.h> #include <stdlib.h> #include "header.h" int main(int argc, char *argv[]) { int s,i,numOfPlayers; struct player *players; printf("Give the number of players: \n"); scanf("%d",&numOfPlayers); players = (struct player *)calloc(numOfPlayers

Proper way to initialize C++ structs

和自甴很熟 提交于 2019-11-26 23:51:41
问题 Our code involves a POD (Plain Old Datastructure) struct (it is a basic c++ struct that has other structs and POD variables in it that needs to get initialized in the beginning.) Based one what I've read, it seems that: myStruct = (MyStruct*)calloc(1, sizeof(MyStruct)); should initialize all the values to zero, as does: myStruct = new MyStruct(); However, when the struct is initialized in the second way, Valgrind later complains "conditional jump or move depends on uninitialised value(s)"