Question

以下代码是为“武器游戏（NOV19B-SC31）”编写的。经过多次尝试，我观察到的问题是代码中的“战斗”功能能够读取但不能更改字符串中的各个字符。

从以下代码中， “ res”值相对于代码中提供的条件没有变化。初始化值，即0000000000，在整个代码中都保持不变。这样每次的答案都是0。

'''

#define Forn(i,N) for(i = 0;i<N;i++)

void battle(char *s1[11],char *s2[11])
{
int i;
Forn(i,10)
{
    if(s1[i]=='1' && s2[i]=='1'){
        s1[i]='0';
        s2[i]='0';
    }
    else if(s1[i]=='1' || s2[i]=='1'){
        s1[i]='0';
        s2[i]='1';
    }
    printf("%s %s\n",s1,s2);
}
}
int main(void) 
{
int t,count=0;
long long int n,i;
scanf("%d",&t);
while(t>0)
{
    scanf("%lld",&n);
    char s[n][11],res[11]="0000000000";
    Forn(i,n)
        scanf("%s",s[i]);

    Forn(i,n)
        battle(s[i],res);

    Forn(i,11)
    {
        if(res[i]=='1')
            count++;
    }
    printf("%d\n",count);
    t--;
}
return 0;
}

'''

示例输入： 1个 3 1110001101 1010101011 0000000011

示例输出： 4

说明：示例情况1：如果第一人称击败第二人，则武器1、3、7和10被销毁。然后，如果第三人击败第一人（现在拥有武器2、5、8和9），则武器9被销毁，获胜者拥有武器2、5、8和10的武器，则意味着还有4件武器。

Answer 1

您将array of pointer作为battle函数的参数

void battle(char *s1[11],char *s2[11])

但是您需要array of characters。

void battle(char s1[11],char s2[11])

如果启用了警告，

warning.c: In function ‘battle’:
warning.c:10: warning: comparison between pointer and integer
warning.c:10: warning: comparison between pointer and integer
warning.c:11: warning: assignment makes pointer from integer without a cast
warning.c:12: warning: assignment makes pointer from integer without a cast
warning.c:14: warning: comparison between pointer and integer
warning.c:14: warning: comparison between pointer and integer
warning.c:15: warning: assignment makes pointer from integer without a cast
warning.c:16: warning: assignment makes pointer from integer without a cast
warning.c:18: warning: format ‘%s’ expects type ‘char *’, but argument 2 has type ‘char **’
warning.c:18: warning: format ‘%s’ expects type ‘char *’, but argument 3 has type ‘char **’
warning.c: In function ‘main’:
warning.c:34: warning: passing argument 1 of ‘battle’ from incompatible pointer type
warning.c:5: note: expected ‘char **’ but argument is of type ‘char *’
warning.c:34: warning: passing argument 2 of ‘battle’ from incompatible pointer type
warning.c:5: note: expected ‘char **’ but argument is of type ‘char *’

为什么字符串字符在执行“战斗”功能时不改变？

1 个答案: