错误：无法将 'char (*)[6]' 转换为 'char**'

Question

So my problem is the following, im trying to pass an array of strings and trying to find a specific one but for some reason the compiler is giving me this error and i dont get it because im clearly passing an array of strings into the function.

program:

#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <string.h>
#include <iostream>

int find_uid(char* users[],char* uid)
{
    for (int i=0;i<32;i++)
    {
        if (strcmp(users[i],uid) == 0)
            return 1;
    }
    return 0;
}

char users[32][6];

int main()
{
    char* user;
    user = new char[6];
    strcpy(user,"10014");
    strcpy(users[6],user);
    printf("Result %d",find_uid(users,"10014"))
}

Answer 1

For starters using strcpy in this code snippet

user = new char[5];
strcpy(user,"10014");

invokes undefined behavior because the dynamically allocated array does not have a space to store the terminating zero character of the string literal "10014" . You need to allocate an array at least of 6 elements

user = new char[6];

The two-dimensional array users declared like

char users[32][5];

used as a function argument is converted implicitly to a pointer to its first element of the type char ( * )[5] .

On the other hand, the corresponding function parameter has the type char ** . These pointer types are not compatible. So the compiler issues an error.

You need to declare the function at least like

int find_uid( char ( *users )[5],char* uid);

Or i is even better to declare the function like

bool find_uid( const char users[][5], size_t n, const char *uid );

Again the second dimension of the array should be increased

char users[32][6];

if you want to use the string literal "10014" in the array.