针对你遇到的编译错误 "cast to pointer from integer of different size [-Werror=int-to-pointer]",我们可以从以下几个方面进行分析和解决: 理解错误信息: 这个错误通常发生在你尝试将一个整数(integer)转换为指针(pointer)类型时,但整数的大小与指针的大小不匹配。在不同的平台和编译器中,整数和指
"<<std::endl;}elseif(std::shared_ptr<Cat>cat=std::dynamic_pointer_cast<Cat>(animal)){std:...
参考博客:http://blog.chinaunix.net/uid-20548989-id-1667467.html 中断处理函数,需要把一个值存在 void * 中传给回调函数,然后在回调函数内部把 int 值从 void * 中取出。 把一个 int 存到 void * 变量需要先把它转换成 long, 这样能避免:warning: cast to pointer from integer of different size; 方法...
最近学习多线程,写Demo程序时,遇到一个编译器告警,就是在用 pthread_join 函数获取线程返回状态值时,出现 -Wpointer-to-int-cast 告警。下面来看一下具体是啥问题 原Demo 程序如下: #include<stdio.h>#include<stdlib.h>#include<assert.h>#include<pthread.h>void*func_1(void*);void*func_2(void...
其实,在编译的时候GCC就已经给出了警告,Wpointer-to-int-cast意思是将指针转换为整型,二者大小不同。但我们大多数时候可能直接会忽略。 代码语言:javascript 代码运行次数:0 运行 AI代码解释 test.c:Infunction‘main’:test.c:13:12:warning:cast from pointer to integerofdifferent size[-Wpointer-to-int-cas...
#include <stdio.h> int main() { int arr[5] = {1, 2, 3, 4, 5}; void* ptr = &arr[2]; int* intPtr = (int*)ptr; printf("Value at pointer: %d\n", *intPtr); return 0; } 避免不必要的类型转换:在设计程序时,尽量避免将指针转换为整数类型,除非你确切知道你在做什么,并且...
/* * Even though sizeof(SOCKET) is 8, it's safe to cast it to int, because * the value constitutes an index in per-process table of limited size * and not a real pointer. And we also depend on fact that all processors * Windows run on happen to be two's-complement, which allo...
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其实,在编译的时候GCC就已经给出了警告,Wpointer-to-int-cast意思是将指针转换为整型,二者大小不同。但我们大多数时候可能直接会忽略。 test.c: In function ‘main’: test.c:13:12: warning: castfrompointer to integer of different size [-Wpointer-to-int-cast]13|intptr = (int)p; ...