我有这段代码
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!
}
}
问题是，当第二次检查1.5时，Runnerm[i][j]从Runners[i+1][j]获取值
简而言之，当我更改值时，之前存储的Runnerm值将获得当前存储的Runners的值。
为什么？我找不到原因..
编辑：“检查1”和“检查1.5”的存在是为了在循环的任何给定时间检查值。所以，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，因为我不再需要检查这些值。谢谢大家回答，但我不明白为什么你建议做一些检查，他们在那里进行调试。 ：\
编辑找到导致问题的原因，当我使用其他版本的程序使用较少的数组时，问题就解决了，当我开始添加随机未使用的数组时，问题出现了。 这是为什么？ （新问题） 可能是内存与Runners [i] [j]使用的内存重叠了吗？
答案 0 :(得分：2)
您的printf语句似乎显示下一个跑步者/下一次，但记录的是跑步者/时间之前：
printf("\nEnter the time of lap %d for runner %d in minutes: ", j+1, i+1);
^^^^ ^^^^
为什么不i j
答案 1 :(得分：1)
移动以下代码行：
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]);
到外部 j循环，删除冗余提示，并删除用作循环边界的魔术常量。
#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]);
}
虽然就个人而言，我认为非功能性是您遇到的最少问题：您的代码被过度注释并浸泡WET，并且您的输出中包含太多换行符。
答案 2 :(得分：1)
将问题代码充实为实际程序：
#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!
添加了以下if(j > ...，以便不打印未初始化的数组值：
if(j > 0) printf("\t %d minutes -- %d seconds \n", Runnerm[i][1], Runners[i][1]); // Incorrect Check!
if(j > 1) printf("\t %d minutes -- %d seconds \n", Runnerm[i][2], Runners[i][2]); // Incorrect Check!
if(j > 2) printf("\t %d minutes -- %d seconds \n", Runnerm[i][3], Runners[i][3]); // Incorrect Check!
}
}
return(0);
}
然后我编译了代码：
SLES11SP2:~/SO> gcc -Wall -o test *.c
并运行代码：
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>
似乎工作正常。