如何编写代码让用户将不同的值输入到数组中

Question

我是 C 的初学者（学习免费资源），我正在尝试编写一个简单的代码来将用户输入的一组数字添加到一组数组中......我做错了什么？

int numbers;
    int number=1;
    int counter = 0;
    int pot[100];
    int calculate;
    int result;

printf("how many numbers do you want to calculate?:"); //asking user how many different integers would be calculated

scanf("%d",&numbers);

while (counter < numbers)

{
    printf("input number %d\n", number);//asking user to input figures to be calculated
    scanf("%d",&pot[0]+1); //user would input values into spaces in the array

    counter++;
    number++;

}


printf("Please press 1 for addition and 2 for multiplication");
scanf("%d",&calculate);

switch (calculate)

{
case 1: result = pot[0]+ pot[0]+1;//this is supposed to add all the individual values within the arrays

printf("the result is %d", result);
break;

case  2: result = pot[0]* pot[0]+1;//this is supposed to multiply all the individual values within the arrays
break;
}



    return 0;

Answer 1

int numbers;
int number=1;
// int counter = 0; instead used `i`
int pot[100];
int calculate;
int result;
int i = 0;

printf("how many numbers do you want to calculate?:"); //asking user how many different integers would be calculated

scanf("%d",&numbers);

printf("input %d numbers \n", numbers);//asking user to input figures to be calculated

for(i = 0; i < numbers; i++)
{
    printf("input number %d \n", number);
    scanf("%d", &pot[i]);   //user would input values into spaces in the array
    number++;
}

//scanf("%d",&pot[0]+1);     /* this line is wrong*/
//counter++; 
//number++;     


printf("Please press 1 for addition and 2 for multiplication");
scanf("%d",&calculate);

switch (calculate)

{
case 1: result = 0;//this is supposed to add all the individual values within the arrays
        for(i = 0;i < numbers; i++)
        {
             result = result + pot[i];
        }

        printf("the result is %d", result);
        break;

case  2: result = 1;//this is supposed to multiply all the individual values within the arrays
         for(i = 0;i < numbers; i++)
         {
             result = result * pot[i];
         }
         printf("the result is %d", result);
         break;
}



return 0;

此代码应该适用于您的程序。

Answer 2

我不确定你到底想做什么，所以这是我的猜测：

int sum(const int *arr, size_t n)
{
    int ret;

    for (ret = 0; --n >= 0; ret += arr[n])
        ;

    return ret;
}
long mpy(const int *arr, size_t n)
{
    long ret;

    for (ret = 1; --n >= 0; ret *= arr[n])
        ;

    return ret;

}
// remove trailing newline
void remTrailingNl(char *buf)
{
    if (buf[strlen(buf) - 1] == '\n')
        buf[strlen(buf) - 1] = '\0';
}

int main()
{
    int *arr, choice, tmp;
    char line[80], *endp;
    size_t n;

    printf("How many numbers are in the array?");
    fgets(line, 80, stdin);
    remTrailingNl(line);
    n = strtol(line, &endp, 10);
    // **endp will be the null character if conversion was successful
    if (**endp == '\0')
        if ((arr = malloc(n * sizeof(int))) == NULL) {
            perror("malloc");
            return 1;
        }

    printf("Begin entering your numbers, each number to a line\n");

    // walking backwards, populate the array
    while (n-- >= 0) {
        fgets(line, 80, stdin);
        remTrailingNl(line);
        tmp = strtol(line, &endp, 10);
        if (**endp == '\0')
            arr[n] = tmp;
        else {
            fprintf(stderr, "Invalid number at position %d\n", n + 1);
            return 1;
        }
    }

    printf("Enter 1 for addition, and 2 for multiplication\n");
    fgets(line, 80, stdin);
    remTrailingNl(line);
    choice = strtol(line, &endp, 10);
    if (**endp == '\0' && choice == 1 || choice == 2) {
        switch (choice) {
            case 1: // add
                printf("Sum is %d\n", sum(arr, n));
                break;
            case 2:
                printf("Product is %ld\n", mpy(arr, n));
                break;
        }
    } else {
        fprintf(stderr, "Invalid choice\n");
        return 1;
    }
    // heap memory must be freed
    free(arr);
    // return 0 on success
    return 0;
}

我们使用long 表示mpy，因为乘法可以给出非常大的答案。另外，不要使用scanf，因为虽然看起来很无辜，但要正确却很棘手。 strtol 更好，因为它可以详细说明号码无效的原因。阅读手册页here。我们在调用fgets 之后立即调用remTrailingNl，因为fgets 总是在缓冲区中放置一个尾随换行符。

Answer 3

如果你想要一个简单的代码来添加数组中的数字，其中数字是由用户给出的，你可以跟着这个。

int i, num, ans=0, pot[100];
printf("Give the number of elements:");
scanf("%d",&num);
printf("Enter numbers");
for(i=0;i<num;i++) {
  scanf("%d",&pot[i]);
  ans = ans + pot[i];
}
printf("Sum is %d", ans);
}

这种方式直接显示添加，而不是一次又一次地扫描它，必须循环并再次添加它，一个一个。

(而且我也是C学习者，如有错误请指正。) 希望这可以简化它。

Answer 4

So, finally got a working program!!

int pot[100];
int numbers;
int count;
int number = 1;
int calculate;
int result;


printf("how many numbers do you want to calculate:");
scanf("%d", &numbers);

for(count=0;count<numbers;count++)

{
printf("enter number %d:\n", number);
scanf("%d",&pot[count]);
number++;

}

printf("Please press 1 for addition and 2 for multiplication");
scanf("%d",&calculate);

switch (calculate)

{
case 1: result = 0;

        for(count = 0;count < numbers; count++)
        {
             result = result + pot[count];
        }

        printf("the result is %d", result);
        break;

case  2: result = 1;

         for(count = 0;count< numbers; count++)
         {
             result = result * pot[count];
         }
         printf("the result is %d", result);
         break;
}

返回 0;