為什么這段代碼不產生輸出？

Question

我正在編寫一個簡單的程序來顯示兩個會聚列車之間的距離/時間。 我想測試程序並通過函數 float 收斂返回輸出值，然后將其應用到主函數，但收斂函數似乎不起作用。

#include <stdio.h>

float converge(int x, int y, int z) {
    return x / (y + z);
}

int main(void) {
    int dist, speed1, speed2;
    printf("Enter distance in miles");
    scanf("%d\n", &dist);
    printf("Enter train 1 speed and train 2 speed in mph");
    scanf("%d%d\n", &speed1, &speed2);

    converge(dist, speed1, speed2);

    return 0;
}

Answer 1

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


float converge (float x, float y, float z)
{
  int time=x/(y+z);
  return time;
}

int main ()
{
    float dist, speed1, speed2;
    printf("Enter distance in miles:\t");
    scanf("%f", &dist);
    printf("Enter speed of first train in mph:\t");
    scanf("%f", &speed1);
    printf("Enter speed of second train in mph:\t");
    scanf("%f", &speed2);

    printf("Time between this two trains is %f",converge(dist, speed1, speed2));


}

Answer 2

為什么這段代碼不產生輸出？

由於提供的代碼中沒有可能導致此輸出的語句，因此它不會從converge()的結果中產生預期輸出的輸出。

例如，您需要在調用converge()之后使用一個printf()語句來打印converge()的結果：

#include <stdio.h>

float converge (int x, int y, int z)
{
  return x/(y+z);
}

int main (void)
{
  int dist, speed1, speed2;
  float converge_result;

  printf("Enter the distance between the two trains in miles:\n");
  scanf("%d", &dist);
  printf("\n");

  printf("Enter the speed of train 1 and the speed of train 2 in mph:\n");
  scanf("%d %d", &speed1,&speed2);
  printf("\n");


  converge_result = converge(dist, speed1, speed2);

  printf("The time until the two trains encounter each other is:\n %f",converge_result);

  return 0;
}

或者：

#include <stdio.h>

float converge (int x, int y, int z)
{
  return x/(y+z);
}

int main (void)
{
  int dist, speed1, speed2;


  printf("Enter the distance between the two trains in miles:\n");
  scanf("%d", &dist);
  printf("\n");

  printf("Enter the speed of train 1 and the speed of train 2 in mph:\n");
  scanf("%d %d", &speed1,&speed2);
  printf("\n");

  printf("The time until the two trains encounter each other is: \n%f ",
  converge(dist,speed1,speed2);

  return 0;
}

順便說一句，時間距離的計算似乎不正確或至少不完整。

Answer 3

您的代碼中有多個問題：

• converge執行整數算術並將結果轉換為float僅用於返回值。 如果你想計算一個小數，你應該把它改成： double converge(int x, int y, int z) { return (double)x / ((double)y + z); } double converge(int x, int y, int z) { return (double)x / ((double)y + z); }或更好的使用double的輸入值和參數類型：

   double converge(double x, double y, double z) { return x / (y + z); }

• scanf()轉換格式中有尾隨換行符：這將導致scanf()消耗在數字之后鍵入的任何尾隨空格，包括在提示中鍵入的任意數量的換行符。 只要您輸入空行，您就不會收到第二個提示。 從格式字符串中刪除這些\\n 。

• 不打印計算結果。

這是一個修改后的版本：

#include <stdio.h>

double converge(double x, double y, double z) {
    return x / (y + z);
}

int main(void) {
    double dist = 0, speed1 = 0, speed2 = 0;

    printf("Enter distance in miles: ");
    scanf("%lf", &dist);

    printf("Enter train 1 speed and train 2 speeds in mph: ");
    scanf("%lf%lf", &speed1, &speed2);

    if (speed1 + speed2 <= 0)
        printf("No collision\n");
    else
        printf("Time until collision: %f seconds\", 3600 * converge(dist, speed1, speed2));

    return 0;
}