@文心快码error: pointer of type 'void *' used in arithmetic 文心快码在C/C++编程中,遇到error: pointer of type 'void *' used in arithmetic这类错误通常意味着你尝试对void *类型的指针进行了算术运算,这是不被允许的。下面我将详细解释这个错误的含义、void指针的特性和限制、常见导致错误的场景、修正...
警告解决方案: 在void*指针变量前加一个(uint8_t *),例如: //command.payload是指向void的指针类型 //修改之前: void* ptr = command.payload + command.len; //修改之后: void* ptr = (uint8_t *)command.payload + command.len;版权声明:本文为MashiMaroJ原创文章,遵循 CC 4.0 BY-SA 版权协议,转载...
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对void* 类型的指针进行加减运算的时候,这是不安全的,因为编译器并不知道你指针的单元是多少,需要移动多长的距离。作为扩展,大部分编译器都允许给void 指针做算术运算。
可以初始化，但是printf那句还需要写成*((int*)a)这样编译器才知道这个地方是需要用整形指针来求值。
编译报错 warning: pointer of type ‘void *’ used in arithmetic [-Wpointer-arith] 编译原因: 使用void*指针进行算术运算。 因为指针运算需要知道指针步长,即指针指向的对象的大小 如上述运算,+1的结果是不一样的,对p+1会前进4个字节,对c+1只会前进1个字节。 警告解决方案: 在void*指针变量前加一个...
指针数组: Type *array[] 数组指针: Type (*array)[]指针函数: Type *function(void)函数指针: Type (*function)(void)函数指针数组: Type (*array[])(void)今天是20200305指针数组,数组指针,指针函数.函数指针,函数指针数组,指针的指针,多级指针你值得拥有 ...
Void pointers can point to any memory chunk. Hence the compiler does not know how many bytes to increment/decrement when we attempt pointer arithmetic on a void pointer. Therefore void pointers must be first typecast to a known type before they can be involved in any pointer arithmetic. ...
#include <iostream> void bar(int& z) { z = 456; } int main(int argc, char** argv) { int x = 123; bar(x); std::cout << x << std::endl; // This prints "456" return 0; } On the other hand, if a variable of reference type is passed into a function, the effect will...
vm.c: In function ‘find_irep_symbol’: vm.c:38:5: error: pointer of type ‘void _’ used in arithmetic [-Wpointer-arith] p += 4; ^ vm.c:41:7: error: pointer of type ‘void *’ used in arithmetic [-Wpointer-arith] p += 2+s+1; // size(2 bytes) ...