如何将值放入const char**？

Question

如果我声明一个变量const char**stringTable,如果它是一个const,我怎么能把值放到它？(它必须是const,因为我应该使用的函数将const char**作为参数.)

编辑:不,你不能隐式地从char**转换为const char**.编译器抱怨:无法将参数3从'char**'转换为'const char**'

Answer 1

哇,我很惊讶没有人得到这个!也许我可以获得死灵法师徽章.隐式const转换仅扫描一层深度.所以a char*可以成为一个const char*但是它不会在一个char**类型内深入挖掘以找到需要改变的东西来使它成为一个const char**.

#include <iostream>
using namespace std;

void print3( const char **three ) {
        for ( int x = 0; x < 3; ++ x ) {
                cerr << three[x];
        }
}

int main() {
         // "three" holds pointers to chars that can't be changed
        const char **three = (const char**) malloc( sizeof( char** ) * 3 );
        char a[5], b[5], c[5]; // strings on the stack can be changed
        strcpy( a, "abc" ); // copy const string into non-const string
        strcpy( b, "def" );
        strcpy( c, "efg" );
        three[0] = a; // ok: we won't change a through three
        three[1] = b; // and the (char*) to (const char*) conversion
        three[2] = c; // is just one level deep
        print3( three ); // print3 gets the type it wants
        cerr << endl;
        return 0;
}