Question

我的函数需要一个void *缓冲区，并且必须用char * buffer_rep内部的内容填充它。但是当我写这篇文章时：

int function(void *buffer, int size)
{
   char buffer_rep[512];
   // my buffer_rep is filled up with the call of another function.
   // It is not empty
   for (int i = 0; i < size; i++) {   // size is below 512
       buffer[i] = buffer_rep[i]; 
   }
return 0;
}

for表达式内的行给我错误“无效使用void表达式”。如果我尝试像这样投射缓冲区：

(char *)buffer[i] = buffer_rep[i];

我收到相同的错误，并显示警告“无效的类型转换”。我也尝试用（char）代替（char *）。我在C语言方面没有太多经验，我尝试自己寻找解决方案，但通常人们在投射时不会出现此错误。我认为有些明显的东西我看不到。

Answer 1

您必须强制转换指针，但是强制转换的问题是您强制转换了 element 而不是指针。即表达式(char *)buffer[i]等于(char *)(buffer[i])。

如果在正确的位置使用括号，应该没问题：((char *)buffer)[i]

Answer 2

您必须将强制转换应用于null: Array(1) [Object] length: 1 __proto__: Array(0) [, …] 0: Object {fantasypoints: 324.15, players: Array(8), salary: 48800} fantasypoints: 324.15 players: Array(8) [Object, Object, Object, …] length: 8 __proto__: Array(0) [, …] 0: Object {_max_exposure: null, _projected_ownership: null, first_name: "Ray", …} 1: Object {_max_exposure: null, _projected_ownership: null, first_name: "Clearlove", …} 2: Object {_max_exposure: null, _projected_ownership: null, first_name: "Scout", …} 3: Object {_max_exposure: null, _projected_ownership: null, first_name: "Hope", …} 4: Object {_max_exposure: null, _projected_ownership: null, first_name: "Meiko", …} 5: Object {_max_exposure: null, _projected_ownership: null, first_name: "SnowFlower", …} 6: Object {_max_exposure: null, _projected_ownership: null, first_name: "Arce", …} 7: Object {_max_exposure: null, _projected_ownership: null, first_name: "Infinity eSports", …} salary: 48800 __proto__: Object {constructor: , __defineGetter__: , __defineSetter__: , …}，而不是buffer。

将代码更改为

buffer[i]

投射void *缓冲区时无效使用void表达式

2 个答案: