如何在一行上打印所有数组项？

Question

我有以下使用数组的代码。 当我运行程序时，它在 9 条不同的行上显示了 9 次结果，但我只想显示一次结果：

#include <stdio.h>

int main(void) {

    int sin_num[9];
    int num1;

    for (num1 = 0; num1 < 9; num1++) {
        printf("Enter your SIN number one by one:");
        scanf("%d", &sin_num);
    }

    for (num1 = 0; num1 < 9; num1++) {
        printf("%d \n", &sin_num[num1]);
    }
}

Answer 1

程序中的 3 个问题：

1. 阅读scanf()手册页

   scanf("%d", &sin_num);` ^^^^^^^ here, sin_num is array of int, so scan should take it into its element and not to its base address.

   用如下所示的索引替换它， scanf("%d", &sin_num[i]);

    for (num1 = 0; num1 < 9; num1++) { printf("Enter your SIN number one by one:"); scanf("%d", &sin_num[i]); }

2. 阅读printf(3)手册页。

    printf("%d \\n", &sin_num[num1]); /* ^, here no need of & as you are looping over array. */ /* Correction => printf("%d \\n", sin_num[num1]); */ for (num1 = 0; num1 < 9; num1++) { printf("%d \\n", sin_num[num1]); }

3. 避免多行

    printf("%d \\n", sin_num[num1]); /* ^^ as per your requirement, you don't need every element on new line so it should be removed. */ for (num1 = 0; num1 < 9; num1++) { printf("%d \\n", sin_num[num1]); }

Answer 2

去掉第二个 for 循环中的“\\n”。

for(num1=0; num1<9; num1++) {
    printf("%d ", sin_num[num1]);
}
print("\n");