C ++的错误：字段'INT []'不完全类型 [英] C++ error: field has incomplete type 'int []'
问题描述
我在C使虚拟机++和我碰到这个错误，
错误：字段具有不完全类型'诠释[]'
诠释instrarr [];
我绝对不知道什么是错的int数组。有人可以来看看，让我知道我做错了，我一直在看它了一个多小时，我似乎无法找到什么微小的细节，我必须离开了。我的整个文件低于柜面你需要它，以供参考。
的#include＆LT;＆iostream的GT;
＃包括LT＆;＆的fstream GT;
＃包括LT＆;串GT;
＃包括LT＆;＆sstream GT;使用命名空间std;类虚拟机{
私人的： 串finfo;
串FILEDATA;
串fconv;
串指令; 诠释instrarr [];
INT zerocount []; 上市： / *辅助函数* / INT countinstrs（字符串s）{
诠释计数= 0; 的for（int i = 0; I＆LT; s.size（）;我++）
如果（S [I] ==''）的计++; 返回计数;
} INT countlzeros（字符串s）{
诠释计数= 0; 的for（int i = 0; I＆LT; s.size（）;我++）
如果（S [I] =='0'）{
算上++;
}其他{
I = s.size（）+ 1;
}
返回计数;
} 字符串load_program（字符串文件）{
ifstream的rdfile（文件）;
而（rdfile＆GT;＆GT;指令）{
FILEDATA + =指令;
FILEDATA + =;
}
rdfile.close（）;
返回FILEDATA;
} 字符串convert_program（字符串fconv）{
INT instrcount = countinstrs（fconv）;
stringstream的HEXTOINT;
无符号整型值;
字符串s = fconv;
字符串分隔符=;
为size_t POS = 0;
字符串标记;
INT I = 0;
而（（POS = s.find（分隔符））！=字符串::非营利组织）{
令牌= s.substr（0，POS）
INT zeroc = countlzeros（令牌）;
// zerocount [I] = zeroc;
stringstream的HEXTOINT（标记）;
HEXTOINT＆GT;＆GT;十六进制＆GT;＆GT;值;
// instrarr [I] =价值;
COUT＆LT;＆LT;值＆LT;＆LT; ENDL;
s.erase（0，位置pos + delimiter.length（））;
我++;
}
返回;
} 无效run_program（字符串文件）{
finfo = load_program（文件）;
fconv = convert_program（finfo）;
// execute_program（）;
}
};INT主（INT ARGC，CHAR *的argv []）{ VM次;
rd.run_program（ARGV [1]）;
返回0;}
这是pretty简单， INT [] 是一个不完整的类型，因为它缺少相关信息有多大是。在函数调用的参数，它进入与声明一个指针，而不是阵列同义，但对于defintions，如在code时，编译器当然需要知道数组多大，以便为它分配存储空间。
I'm making a virtual machine in C++ and I've run into this error,
error: field has incomplete type 'int []'
int instrarr[];
I have absolutely no idea what is wrong with the int array. Can someone take a look and let me know what I've done wrong, I've been looking at it for over an hour and I can't seem to find what tiny detail I must have left out. My entire file is below incase you need it for reference.
#include <iostream>
#include <fstream>
#include <string>
#include <sstream>
using namespace std;
class vm {
private:
string finfo;
string filedata;
string fconv;
string instruction;
int instrarr[];
int zerocount[];
public:
/* Helper functions */
int countinstrs(string s) {
int count = 0;
for (int i = 0; i < s.size(); i++)
if (s[i] == ',') count++;
return count;
}
int countlzeros(string s) {
int count = 0;
for (int i = 0; i < s.size(); i++)
if (s[i] == '0') {
count++;
} else {
i = s.size() + 1;
}
return count;
}
string load_program(string file) {
ifstream rdfile(file);
while(rdfile >> instruction) {
filedata += instruction;
filedata += ",";
}
rdfile.close();
return filedata;
}
string convert_program(string fconv) {
int instrcount = countinstrs(fconv);
stringstream hextoint;
unsigned int value;
string s = fconv;
string delimiter = ",";
size_t pos = 0;
string token;
int i = 0;
while ((pos = s.find(delimiter)) != string::npos) {
token = s.substr(0, pos);
int zeroc = countlzeros(token);
//zerocount[i] = zeroc;
stringstream hextoint(token);
hextoint >> hex >> value;
//instrarr[i] = value;
cout << value << endl;
s.erase(0, pos + delimiter.length());
i++;
}
return "";
}
void run_program(string file) {
finfo = load_program(file);
fconv = convert_program(finfo);
//execute_program();
}
};
int main(int argc, char* argv[]) {
vm rd;
rd.run_program(argv[1]);
return 0;
}
It's pretty simple, int[] is an incomplete type, as it lacks information about how large it is. In function call parameters, it goes synonymous with declaring a pointer instead of an array, but for defintions, as in your code, the compiler certainly needs to know how large the array is, in order to allocate storage for it.
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