C语言的一个关于类型的小陷阱

类型的差别造成奇怪的运行结果——输出错误、无法继续输入等

新增：

找出下面代码的错误：

# include <stdio.h>
# include <string.h>
 
# define MAXN 10005
 
int main()
{
    short m, n, i, ans, tmp, t[MAXN];
 
    while (~scanf("%d%d", &n, &m))
    {
        memset(t, 0, sizeof(t));
        for (i = 1; i <= m; ++i)
        {
            scanf("%d", &tmp);
            t[tmp] = t[tmp]+1;
        }
        ans = tmp = 0;
        for (i = 1; i <= n; ++i)
            if (t[i] > tmp)
            {
                tmp = t[i];
                ans = i;
            }
        printf("%d\n", ans);
    }
 
    return 0;
}

题一：输入两个实数（0~1000），输出它们的和，并保留四位小数（多组输入）。

# include <stdio.h>
int main()
{
    double a,b;
    while (scanf("%f%f", &a,&b))
        printf("%.4f\n", a+b);
    return 0;
}

# include <stdio.h>
int main()
{
    double a,b;
    while (scanf("%lf%lf", &a,&b))
        printf("%.4lf\n", a+b);
    return 0;
}

题二：输入两个整数(0~1000)，输出它们的和（多组输入）。

# include <stdio.h>
int main()
{
    short int a,b;
    while (scanf("%d%d", &a,&b))
        printf("%d\n", a+b);
    return 0;
}

# include <stdio.h>
int main()
{
    int a,b;
    while (scanf("%d%d", &a,&b))
        printf("%d\n", a+b);
    return 0;
}

找出下列代码错误之处：

// 1003 UC Browser
# include <stdio.h>
# define MAXN 105
short int expr[] = {0,1,3,6,10,15};
int main()
{
    short int T, n, tot, cnt, i;
    char a[MAXN];
    scanf("%d", &T);
    while (T > 0)
    {
        scanf("%d%s", &n,a);
        tot = 0;
        for (i=0; i<n; ++i)
        {
            cnt = 0;
            while ('1'==a[i]){++cnt;++i;}
            tot += (cnt/5)*15+expr[cnt%5];
        }
        printf("%d\n",(tot<75 ? (tot+5)/10:8));
        --T;
    }
    return 0;
}

正确的在这里：

// 1003 UC Browser
# include <stdio.h>
# define MAXN 105
short int expr[] = {0,1,3,6,10,15};
int main()
{
    int T, n, tot, cnt, i;         // short int 改为 int
    char a[MAXN];
    scanf("%d", &T);
    while (T > 0)
    {
        scanf("%d%s", &n,a);
        tot = 0;
        for (i=0; i<n; ++i)
        {
            cnt = 0;
            while ('1'==a[i]){++cnt;++i;}
            tot += (cnt/5)*15+expr[cnt%5];
        }
        printf("%d\n",(tot<75 ? (tot+5)/10:8));
        --T;
    }
    return 0;
}