你的本意是把i传到线程函数里，相信你传进去也不会做指针使用。因此从你的角度上来讲，这没有错。但...
char*ptr;//此后省略部分代码if( (int)ptr==-1)//出错地方 那句话的意思是从 void* 到 int 的转换丢失精度,相信看到解释有些人就明白了, 此问题只会出现在X64位的Linux上,因为在64位的机器上指针占用8个字节,int 占用四个字节,所以才会出现这样的问题, 解决方法: (long)ptr == -1 就好了...
代码如下: for (i = 0; i < nthreads; i++) { nodes[i] = (int**)realloc(nodes[i], (sizeof(int *)*lnpages[i])); pthread_create(&threads[i], NULL, worker, (void *)i); } 编译器报错是 (void *)i 处,错误说明是 Error: cast to 'void *' from smaller integer type 'int' ...
I have avoid* pointer. I want to cast the value at that pointer to an int and add it the value at anothervoid* x. I have already tried: int But this gives me aoperand of type 'void' where arithmetic or pointer type is requirederror You can't dereference a pointer to void. That ...
classCMyClass{public:explicitCMyClass(intiBar)throw(){ }staticCMyClassget_c2(); };intmain(){ CMyClass myclass =2;// C2440// try one of the following// CMyClass myclass{2};// CMyClass myclass(2);int*i;floatj; j = (float)i;// C2440, cannot cast from pointer to int to ...
I wonder why is it necessary to cast a void pointer to an int * or char * before printing the contents of the address in memory, even though we tell the printf() function how to interpret the data in memory? Let's say that we have the following code: in...
C 指针的小小实验 更新: 空白指针,也被称为通用指针,是一种特殊类型的指针,可以指向任何数据类型的对象! 空白指针像普通指针一样被声明,使用void关键字作为指针的类型。 The void pointer, also known as the…
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int i; for (i = 0; i < n; i++) { arr[i] = 0; } } int main() { int grades[50]; setToZero(grades, 50); return 0; } In this program, the setToZero function takes a pointer to an integer as its first parameter. When we call it with “setToZero(grades, 50)”, the...
{chara[20];int*p=new(a)int;return0; } 深入理解LINUX虚拟内存管理 #include<stdio.h>intadd(inta,intb) {returna+b; }intman() {//int (*fun)(int)=(int(*)(int))add;typedefint(*fun)(int); fun f=(fun)add;intr=fun(20);