C中的循环出错了

我有这个密码

for (i = 0; i < s; i++) // for i from 0 to Runners
{
    for (j = 0; j < 4; j++) //for j from 0 to laps
    {
        printf("\nEnter the time of lap %d for runner %d in minutes: ", j+1, i+1); // prompt for time for each runner in minutes
        while (scanf("%d", &Runnerm[i][j]) != 1) // While scanf returns not equal to 1 (value parsed == TRUE)
        {
            while ((ch = getchar()) != '\n') putchar(ch); // check if ch=character, compare last entry (\n if character)
            printf(" is not an integer.\nPlease enter only an "); // print error
            printf("integer, such as 1, 5, or 9 : \n"); // cnt print error
        }

        printf("Enter the time of lap %d for runner %d in seconds: ", j+1, i+1); //prompt for time for each runner in seconds
        while (scanf("%d", &Runners[i][j]) != 1) // While scanf returns not equal to 1 (value parsed == TRUE)
        {
            while ((ch = getchar()) != '\n') putchar(ch); // check if ch=character, compare last entry (\n if character)
            printf(" is not an integer.\nPlease enter only an "); // print error
            printf("integer, such as 1, 5, or 9 : \n"); // cnt print error
        }

        printf(" \n check 1 \n ");
        printf("\n %d minutes -- %d seconds \n", Runnerm[i][j], Runners[i][j]); // Correct Check!

        printf(" \n check 1.5 \n ");
        printf("\n %d minutes -- %d seconds \n", Runnerm[i][0], Runners[i][0]); // Incorrect Check!
        printf("\n %d minutes -- %d seconds \n", Runnerm[i][1], Runners[i][1]); // Incorrect Check!
        printf("\n %d minutes -- %d seconds \n", Runnerm[i][2], Runners[i][2]); // Incorrect Check!
        printf("\n %d minutes -- %d seconds \n", Runnerm[i][3], Runners[i][3]); // Incorrect Check!

    }
}

（i=0；ifor//i从0到 { 对于（j=0；j<4；j++）//对于从0到圈的j { printf（“\n以分钟为单位输入跑步者%d圈的时间：”，j+1，i+1）；//以分钟为单位提示每个跑步者的时间 while（scanf（“%d”，&Runnerm[i][j]）！=1）//而scanf返回的值不等于1（解析后的值==TRUE） { while（（ch=getchar（））！='\n'）putchar（ch）；//检查ch=character，比较最后一个条目（\n如果character） printf（“不是整数。\n请只输入一个”）；//打印错误 printf（“整数，如1、5或9:\n”）；//cnt打印错误 } printf（“以秒为单位输入跑步者%d圈的时间：”，j+1，i+1）；//以秒为单位提示每个跑步者的时间 while（scanf（“%d”，&Runners[i][j]）！=1）//而scanf返回的值不等于1（解析后的值==TRUE） { while（（ch=getchar（））！='\n'）putchar（ch）；//检查ch=character，比较最后一个条目（\n如果character） printf（“不是整数。\n请只输入一个”）；//打印错误 printf（“整数，如1、5或9:\n”）；//cnt打印错误 } printf（“\n检查1\n”）； printf（“\n%d分钟--%d秒\n”，Runnerm[i][j]，Runnerm[i][j]）；//检查正确！ printf（“\n检查1.5\n”）； printf（“\n%d分钟--%d秒\n”，Runnerm[i][0]，Runnerm[i][0]）；//检查不正确！ printf（“\n%d分钟--%d秒\n”，Runnerm[i][1]，Runnerm[i][1]）；//检查不正确！ printf（“\n%d分钟--%d秒\n”，Runnerm[i][2]，Runnerm[i][2]）；//检查不正确！ printf（“\n%d分钟--%d秒\n”，Runnerm[i][3]，Runnerm[i][3]）；//检查不正确！ } } 问题是，当check 1.5第二次运行时，

Runnerm[i][j]

从

runner[i+1][j]

简而言之，当我更改值时，

Runnerm

的先前存储值将获得当前存储的运行程序的值

为什么呢？我找不到原因

编辑：存在“检查1”和“检查1.5”的全部目的是检查循环任何给定时间的值。因此，即使我删除它们，程序仍然会被破坏。 此外，当这一切都修复后，检查将被删除，因为我将不再需要检查值。谢谢大家的回答，但我不明白你们为什么建议对这些检查做些什么，它们是用来调试的\

Edit找到了问题的原因，当我使用使用较少数组的程序的另一个版本时，问题得到了解决，当我开始添加随机未使用的数组时，问题出现了。 为什么呢？（新问题）

---

是否内存与跑步者[i][j]使用的内存重叠？

您的

printf

语句显示的是下一个跑步者/下一次，但记录的是跑步者/下一次：

printf("\nEnter the time of lap %d for runner %d in minutes: ", j+1, i+1);
                                                                ^^^^ ^^^^

---

为什么不

i

j

您的

printf

语句似乎正在显示下一个跑步者/下一次，但正在记录之前的跑步者/时间：

printf("\nEnter the time of lap %d for runner %d in minutes: ", j+1, i+1);
                                                                ^^^^ ^^^^

---

为什么不移动这些代码行呢

printf(" \n check 1.5 \n ");
printf("\n %d minutes -- %d seconds \n", Runnerm[i][0], Runners[i][0]);
printf("\n %d minutes -- %d seconds \n", Runnerm[i][1], Runners[i][1]);
printf("\n %d minutes -- %d seconds \n", Runnerm[i][2], Runners[i][2]);
printf("\n %d minutes -- %d seconds \n", Runnerm[i][3], Runners[i][3]);

要在

j

循环之外，请删除冗余提示，并删除用作循环边界的神奇常量

#define RUNNERS s
#define LAPS 4

for (i = 0; i < RUNNERS; i++)
{
    for (j = 0; j < LAPS; j++)
    {
        printf("Enter the time of lap %d for runner %d in seconds: ", j+1, i+1)
        while (scanf("%d", &Runners[i][j]) != 1)
        {
            while ((ch = getchar()) != '\n') putchar(ch);
            printf(" is not an integer.\nPlease enter only an integer, such as 1, 5, or 9 : \n");
        }

        printf(" \n check 1 \n ");
        printf("\n %d minutes -- %d seconds \n", Runnerm[i][j], Runners[i][j]);

    }

    printf(" \n check 1.5 \n ");

    for (j = 0; j < LAPS; j++)
        printf("\n %d minutes -- %d seconds \n", Runnerm[i][j], Runners[i][j]);
}

#定义跑步者
#定义第4圈
对于（i=0；i<0；i++）
{
对于（j=0；j

---

尽管就个人而言，我认为非功能性是您的问题中最小的一个：您的代码被过度压缩和浸泡，并且您的输出中有太多的换行符。

移动这些代码行：

printf(" \n check 1.5 \n ");
printf("\n %d minutes -- %d seconds \n", Runnerm[i][0], Runners[i][0]);
printf("\n %d minutes -- %d seconds \n", Runnerm[i][1], Runners[i][1]);
printf("\n %d minutes -- %d seconds \n", Runnerm[i][2], Runners[i][2]);
printf("\n %d minutes -- %d seconds \n", Runnerm[i][3], Runners[i][3]);

要在

j

循环之外，请删除冗余提示，并删除用作循环边界的神奇常量

#define RUNNERS s
#define LAPS 4

for (i = 0; i < RUNNERS; i++)
{
    for (j = 0; j < LAPS; j++)
    {
        printf("Enter the time of lap %d for runner %d in seconds: ", j+1, i+1)
        while (scanf("%d", &Runners[i][j]) != 1)
        {
            while ((ch = getchar()) != '\n') putchar(ch);
            printf(" is not an integer.\nPlease enter only an integer, such as 1, 5, or 9 : \n");
        }

        printf(" \n check 1 \n ");
        printf("\n %d minutes -- %d seconds \n", Runnerm[i][j], Runners[i][j]);

    }

    printf(" \n check 1.5 \n ");

    for (j = 0; j < LAPS; j++)
        printf("\n %d minutes -- %d seconds \n", Runnerm[i][j], Runners[i][j]);
}

#定义跑步者
#定义第4圈
对于（i=0；i<0；i++）
{
对于（j=0；j

---

尽管就我个人而言，我认为非功能性是你的问题中最小的：你的代码被过度压缩和浸泡，并且你的输出中有太多的换行符。

将问题代码充实到实际的程序中：

#include <stdio.h>

int main()
  {
  int s = 3;
  int i;
  int j;
  int ch;
  int Runnerm[s][4]; //Runners / minutes
  int Runners[s][4]; //Runners / seconds

  for (i = 0; i < s; i++) // for i from 0 to Runners
     {
     for (j = 0; j < 4; j++) //for j from 0 to laps
        {
        printf("\nEnter the time of lap %d for runner %d in minutes: ", j+1, i+1); // prompt for time for each runner in minutes
        while (scanf("%d", &Runnerm[i][j]) != 1) // While scanf returns not equal to 1 (value parsed == TRUE)
           {
           while ((ch = getchar()) != '\n') putchar(ch); // check if ch=character, compare last entry (\n if character)
           printf(" is not an integer.\nPlease enter only an "); // print error
           printf("integer, such as 1, 5, or 9 : \n"); // cnt print error
           }

        printf("Enter the time of lap %d for runner %d in seconds: ", j+1, i+1); //prompt for time for each runner in seconds
        while (scanf("%d", &Runners[i][j]) != 1) // While scanf returns not equal to 1 (value parsed == TRUE)
           {
           while ((ch = getchar()) != '\n') putchar(ch); // check if ch=character, compare last entry (\n if character)
           printf(" is not an integer.\nPlease enter only an "); // print error
           printf("integer, such as 1, 5, or 9 : \n"); // cnt print error
           }

        printf("check 1 \n ");
        printf("\t %d minutes -- %d seconds \n", Runnerm[i][j], Runners[i][j]); // Correct Check!

        printf("check 1.5 \n ");
        printf("\t %d minutes -- %d seconds \n", Runnerm[i][0], Runners[i][0]); // Incorrect Check!

然后我编译了代码：

SLES11SP2:~/SO> gcc -Wall -o test *.c

然后运行代码：

SLES11SP2:~/SO> ./test

Enter the time of lap 1 for runner 1 in minutes: 1
Enter the time of lap 1 for runner 1 in seconds: 2
check 1
   1 minutes -- 2 seconds
check 1.5
   1 minutes -- 2 seconds

Enter the time of lap 2 for runner 1 in minutes: 3
Enter the time of lap 2 for runner 1 in seconds: 4
check 1
   3 minutes -- 4 seconds
check 1.5
   1 minutes -- 2 seconds
   3 minutes -- 4 seconds

Enter the time of lap 3 for runner 1 in minutes: 5
Enter the time of lap 3 for runner 1 in seconds: 6
check 1
   5 minutes -- 6 seconds
check 1.5
   1 minutes -- 2 seconds
   3 minutes -- 4 seconds
   5 minutes -- 6 seconds

Enter the time of lap 4 for runner 1 in minutes: 7
Enter the time of lap 4 for runner 1 in seconds: 8
check 1
   7 minutes -- 8 seconds
check 1.5
   1 minutes -- 2 seconds
   3 minutes -- 4 seconds
   5 minutes -- 6 seconds
   7 minutes -- 8 seconds

Enter the time of lap 1 for runner 2 in minutes: 9
Enter the time of lap 1 for runner 2 in seconds: 10
check 1
   9 minutes -- 10 seconds
check 1.5
   9 minutes -- 10 seconds

Enter the time of lap 2 for runner 2 in minutes: 11
Enter the time of lap 2 for runner 2 in seconds: 12
check 1
   11 minutes -- 12 seconds
check 1.5
   9 minutes -- 10 seconds
   11 minutes -- 12 seconds

Enter the time of lap 3 for runner 2 in minutes: 13
Enter the time of lap 3 for runner 2 in seconds: 14
check 1
   13 minutes -- 14 seconds
check 1.5
   9 minutes -- 10 seconds
   11 minutes -- 12 seconds
   13 minutes -- 14 seconds

Enter the time of lap 4 for runner 2 in minutes: 15
Enter the time of lap 4 for runner 2 in seconds: 16
check 1
   15 minutes -- 16 seconds
check 1.5
   9 minutes -- 10 seconds
   11 minutes -- 12 seconds
   13 minutes -- 14 seconds
   15 minutes -- 16 seconds

Enter the time of lap 1 for runner 3 in minutes: 17
Enter the time of lap 1 for runner 3 in seconds: 18
check 1
   17 minutes -- 18 seconds
check 1.5
   17 minutes -- 18 seconds

Enter the time of lap 2 for runner 3 in minutes: 19
Enter the time of lap 2 for runner 3 in seconds: 20
check 1
   19 minutes -- 20 seconds
check 1.5
   17 minutes -- 18 seconds
   19 minutes -- 20 seconds

Enter the time of lap 3 for runner 3 in minutes: 21
Enter the time of lap 3 for runner 3 in seconds: 22
check 1
   21 minutes -- 22 seconds
check 1.5
   17 minutes -- 18 seconds
   19 minutes -- 20 seconds
   21 minutes -- 22 seconds

Enter the time of lap 4 for runner 3 in minutes: 23
Enter the time of lap 4 for runner 3 in seconds: 24
check 1
   23 minutes -- 24 seconds
check 1.5
   17 minutes -- 18 seconds
   19 minutes -- 20 seconds
   21 minutes -- 22 seconds
   23 minutes -- 24 seconds
SLES11SP2:~/SO> 

---

它似乎工作正常。

将问题代码充实到实际程序中：

#include <stdio.h>

int main()
  {
  int s = 3;
  int i;
  int j;
  int ch;
  int Runnerm[s][4]; //Runners / minutes
  int Runners[s][4]; //Runners / seconds

  for (i = 0; i < s; i++) // for i from 0 to Runners
     {
     for (j = 0; j < 4; j++) //for j from 0 to laps
        {
        printf("\nEnter the time of lap %d for runner %d in minutes: ", j+1, i+1); // prompt for time for each runner in minutes
        while (scanf("%d", &Runnerm[i][j]) != 1) // While scanf returns not equal to 1 (value parsed == TRUE)
           {
           while ((ch = getchar()) != '\n') putchar(ch); // check if ch=character, compare last entry (\n if character)
           printf(" is not an integer.\nPlease enter only an "); // print error
           printf("integer, such as 1, 5, or 9 : \n"); // cnt print error
           }

        printf("Enter the time of lap %d for runner %d in seconds: ", j+1, i+1); //prompt for time for each runner in seconds
        while (scanf("%d", &Runners[i][j]) != 1) // While scanf returns not equal to 1 (value parsed == TRUE)
           {
           while ((ch = getchar()) != '\n') putchar(ch); // check if ch=character, compare last entry (\n if character)
           printf(" is not an integer.\nPlease enter only an "); // print error
           printf("integer, such as 1, 5, or 9 : \n"); // cnt print error
           }

        printf("check 1 \n ");
        printf("\t %d minutes -- %d seconds \n", Runnerm[i][j], Runners[i][j]); // Correct Check!

        printf("check 1.5 \n ");
        printf("\t %d minutes -- %d seconds \n", Runnerm[i][0], Runners[i][0]); // Incorrect Check!

然后我编译了代码：

SLES11SP2:~/SO> gcc -Wall -o test *.c

然后运行代码：

SLES11SP2:~/SO> ./test

Enter the time of lap 1 for runner 1 in minutes: 1
Enter the time of lap 1 for runner 1 in seconds: 2
check 1
   1 minutes -- 2 seconds
check 1.5
   1 minutes -- 2 seconds

Enter the time of lap 2 for runner 1 in minutes: 3
Enter the time of lap 2 for runner 1 in seconds: 4
check 1
   3 minutes -- 4 seconds
check 1.5
   1 minutes -- 2 seconds
   3 minutes -- 4 seconds

Enter the time of lap 3 for runner 1 in minutes: 5
Enter the time of lap 3 for runner 1 in seconds: 6
check 1
   5 minutes -- 6 seconds
check 1.5
   1 minutes -- 2 seconds
   3 minutes -- 4 seconds
   5 minutes -- 6 seconds

Enter the time of lap 4 for runner 1 in minutes: 7
Enter the time of lap 4 for runner 1 in seconds: 8
check 1
   7 minutes -- 8 seconds
check 1.5
   1 minutes -- 2 seconds
   3 minutes -- 4 seconds
   5 minutes -- 6 seconds
   7 minutes -- 8 seconds

Enter the time of lap 1 for runner 2 in minutes: 9
Enter the time of lap 1 for runner 2 in seconds: 10
check 1
   9 minutes -- 10 seconds
check 1.5
   9 minutes -- 10 seconds

Enter the time of lap 2 for runner 2 in minutes: 11
Enter the time of lap 2 for runner 2 in seconds: 12
check 1
   11 minutes -- 12 seconds
check 1.5
   9 minutes -- 10 seconds
   11 minutes -- 12 seconds

Enter the time of lap 3 for runner 2 in minutes: 13
Enter the time of lap 3 for runner 2 in seconds: 14
check 1
   13 minutes -- 14 seconds
check 1.5
   9 minutes -- 10 seconds
   11 minutes -- 12 seconds
   13 minutes -- 14 seconds

Enter the time of lap 4 for runner 2 in minutes: 15
Enter the time of lap 4 for runner 2 in seconds: 16
check 1
   15 minutes -- 16 seconds
check 1.5
   9 minutes -- 10 seconds
   11 minutes -- 12 seconds
   13 minutes -- 14 seconds
   15 minutes -- 16 seconds

Enter the time of lap 1 for runner 3 in minutes: 17
Enter the time of lap 1 for runner 3 in seconds: 18
check 1
   17 minutes -- 18 seconds
check 1.5
   17 minutes -- 18 seconds

Enter the time of lap 2 for runner 3 in minutes: 19
Enter the time of lap 2 for runner 3 in seconds: 20
check 1
   19 minutes -- 20 seconds
check 1.5
   17 minutes -- 18 seconds
   19 minutes -- 20 seconds

Enter the time of lap 3 for runner 3 in minutes: 21
Enter the time of lap 3 for runner 3 in seconds: 22
check 1
   21 minutes -- 22 seconds
check 1.5
   17 minutes -- 18 seconds
   19 minutes -- 20 seconds
   21 minutes -- 22 seconds

Enter the time of lap 4 for runner 3 in minutes: 23
Enter the time of lap 4 for runner 3 in seconds: 24
check 1
   23 minutes -- 24 seconds
check 1.5
   17 minutes -- 18 seconds
   19 minutes -- 20 seconds
   21 minutes -- 22 seconds
   23 minutes -- 24 seconds
SLES11SP2:~/SO> 

---

它似乎工作正常。

数组是如何定义的？int Runnerm[s][4]//跑步者(分钟)[s][4];//跑步者/秒你是说从

检查1

到

检查1.5

，

i

会改变值，也就是说，它会增加

1

？你想在

检查1.5

中看到什么？您将输出大量的UNA