C中的char数组？

Question

我在修改C中的数组时遇到一些问题.我有两个数组,c1和c2.

我想c2通过输入单词array c1contains 手动填充数组.用户键入一个字母,如果在c1中找到此字母,程序会将此字母修复在c2中的适当位置.如果该字母不属于c1,则不执行任何操作.这就像一个"猜词"游戏.这是一个草图:

c1 = MARRON

c2: ******
Type a letter: E


c2: ******
Type a letter: A


c2: *A****
Type a letter: O


c2: *A**O*
Type a letter: P


c2:*A**O*
Type a letter: M


c2: MA**O*
Type a letter: N


c2: MA**ON
Type a letter: R

c2: MARRON

You win.

我的代码并没有真正遵循上面的草图.我的问题是在合适的时间打印星号和字母.我还发现程序用'\0'星号替换了最后一个字符,即使我保留了一个条件来防止这种情况.这是'我的代码:

void complete( char *c2, char c1[], int letter ) {
    int i = 0;
    for( i = 0; i < strlen( c1 ) - 1; i++ )
        if( c1[ i ] == '\0' ) {
            c2[ i ] = '\0';
        }
        else if( letter != c1[ i ] && c2[ i ] != '*' );



        else if ( letter == c1[ i ] ) {
            c2[ i ] = letter;
            c2[ i + 1 ] = '\0';  
        }
        else if (letter != word[ i ] )
            c2[ i ] = '*';
    c2[ i + 1 ] = '\0'; 
}

int main(){
    char c1[] = "ORANGE";
    int length = strlen(c1);

    char *c2 = NULL;
    c2 = malloc(length * (sizeof(c1));

    while( 1 ){
        printf( "What is the secret word?:  ");
        printf( "%s\n", c2 );
        printf( "Propose a letter:  ");
        letter = lirecaracter();//it just reads a character

        char *suite = NULL; 
        suite = strchr(c1, letter );
        if( suite != NULL ){
            complete( c2, c1, letter );
        }

        if( strcmp( c1, c2 ) == 0 ){
            printf( "\n You won, congrats\n" );
            exit( 0 );
        }

    }

}

谢谢.

Answer 1

伪代码:

  WordGuessGame(target[1..n], buffer[1..n], input[1..m])
   1. for i := 1 to n do
   2.    buffer[i] := '*'
   3. print buffer
   4. for i := 1 to m do
   5.    for j = 1 to n do
   6.       if target[j] = input[i] then
   7.          buffer[j] := input[i]
   8.    if buffer = target then
   9.       print "You win!"
  10.       return
  11.    else print buffer
  12. print "You lose..."