C-遍历通过void *的结构数组迭代

问题描述

我有一个函数

struct Analysis reduce (int n, void* results)

其中n是要分析的文件数，我正在将一系列Analysis结构传递给结果。

Where n is the number of files to be analyzed, and I'm passing an array of Analysis structs to results.

分析结构定义如下：

struct Analysis {
  int ascii[128]; //frequency of ascii characters in the file
  int lineLength; //longest line in the file
  int lineNum; //line number of longest line
  char* filename;
}

我已经将空缺*这样，

struct Analysis resArray[n];
struct Analysis* ptr = results;
resArray[0] = ptr[0];

但我不知道如何正确遍历resArray。我已经尝试

but I can't figure out how to iterate through the resArray properly. I've tried

for (i = 0; i < n; i++){
  printf("lineLength: %d\n", resArray[i].lineLength);
}

n = 3，而我得到的是垃圾值。 resArray [0]是正确的，但是resArray [1]是一个非常高的数字，而resArray [2]只是0。为什么resArray [1]或resArray [2]不能提供正确的值？如果我错误地增加了地址，那是有道理的，但是我只是在某个索引处访问数组。

with n = 3, and I'm getting garbage values. resArray[0] is correct, but resArray[1] is an insanely high number and resArray[2] is just 0. Why wouldn't resArray[1] or resArray[2] give the correct values? If I was incrementing the address incorrectly then it would make sense but I'm just accessing the array at a certain index. Pretty lost here!

推荐答案

您不能直接将指针分配给数组，因为它们是不同的类型因为array [n]是结构分析（*）[n]类型，而ptr是结构分析（*）类型。在中查看更多信息。

You can't assign a pointer to an array directly because they are different typessince array[n] is type struct analysis(*)[n] and ptr is type struct analysis(*). Check here for more info.

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