我有一个主文件和一个头文件。
在主文件中，我想从头文件的char函数返回2D char数组。我的char函数如下：
char character_distribution(int length, char redistribution[length][2])
{
char *buffer, distribution[256][2] = {0};
long lSize;
struct Bar result = funct();
buffer = result.x;
lSize = result.y;
length = collect_character_distribution(buffer, lSize, distribution);
reorganize_character_distribution(length, distribution, redistribution);
return redistribution;
}
我的主要功能如下：
#include <stdio.h>
#include "character_distribution.h"
void main()
{
int length;
char distribution[length][2];
distribution = character_distribution(length, distribution[length][2]);
int a;
for(a = 0; a < length; a++)
{
printf("%c\n", distribution[a][0]);
}
}
运行代码时，出现以下错误：
warning: return makes integer from pointer without a cast
我该如何解决该问题？
void character_distribution(int length, char redistribution[][2])
{
char *buffer, distribution[256][2] = {0};
long lSize;
struct Bar result = funct();
buffer = result.x;
lSize = result.y;
length = collect_character_distribution(buffer, lSize, distribution);
reorganize_character_distribution(length, distribution, redistribution);
}
int main()
{
int length = 2; //initialize
char distribution[length][2];
character_distribution(length, distribution);
int a;
for(a = 0; a < length; a++)
{
printf("%c\n", distribution[a][0]);
}
return 0;
}
如果您真的必须返回2d数组，则一种方法（简便方法）是将其放在结构中
struct distribution_struct {
char x[256];
char y[2];
};
struct distribution_struct character_distribution(int length, char redistribution[][2]) {
struct distribution_struct dis;
//initialize the struct with values
//return the struct
}
另一种方法是为函数中的2d数组手动分配内存并返回
char** character_distribution(int length, char redistribution[][2]) {
//use malloc to create the array and a for loop to populate it
}
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