Question

我正在尝试阅读n，然后在n后面会有提示扫描数字和字符串，即“7 dwarf”就像棒球名称一样，你知道“ 30 Rodriguez“当然，正如您在我的代码中看到的那样，它会打印，然后提示另一次scanf，持续n次。： -

int main (void) {



int n, number, i;
char word[1000];

scanf("%d", &n);

for (i=0; i<n; i++) {
    scanf("%d %s", &number, &word);

    printf("Player %d's record: %s\n", number, word);
    printf("Player %d's batting average is\n", number);
}   


system("pause");
return 0;

}

我的输出是......

2
12 harambe
Player 12's record: harambe
Player 12's batting average is
13 Muhammad
Player 13's record: Muhammad
Player 13's batting average is
Press any key to continue . .

我正努力进入： -

2
12 harambe
13 Muhammad

player 12's record: harambe
player 12's batting average is

player 13's record: Muhammad
player 13's batting average is
press any key to continue...

所以在{i} scanf我的输入之后，所有同时出现在一个printf语句中的两个玩家的记录，这个printf将随着n的上升而扩展，在这种情况下它只有2，但是如果我投入5，它应该一次打印5个不同的玩家

Answer 1

您可以根据输入malloc n动态创建数组。

像这样

#include <stdio.h>
#include <stdlib.h>

typedef struct player {
    int number;
    char name[64];
} Player;

int main(void){
    int n;

    scanf("%d", &n);
    Player *recs = malloc(n * sizeof(Player));
    if(recs == NULL){
        perror("malloc");
        exit(EXIT_FAILURE);
    }

    for(int i = 0; i < n; i++) {
        if(2 != scanf("%d %63[^\n]%*c", &recs[i].number, recs[i].name)){
            printf("invalid input.\ninput again!\n");
            while(getchar() != '\n'); //clear input
            --i;
        }
    }
    puts("");
    for(int i = 0; i < n; i++) {
        printf("Player %d's record: %s\n", recs[i].number, recs[i].name);
        printf("Player %d's batting average is \n", recs[i].number);
    }
    free(recs);
    return 0;
}

扫描两个不同的字符和整数需要一次打印

1 个答案: