C 我的程序没有'；t输出在数组中搜索时输入的第一个值。为什么？

例如，如果我为数组写入1,2,3,4,5，然后再次写入1,2,3,4,5以在数组中搜索，它不会输出1。。。我很困惑。你能帮我吗

#include <stdio.h>
#include <stdlib.h>
int main() {
  int i, b, c, n, a;
  int ara[200];
  int number[200];

  printf("Please enter how many numbers you wanna input: \n");
  scanf("%d", &n);
  if (n > 200) {
    // i limited the numbers users can input.
    printf("You can't write higher than 200. It's out of range. ");
  } else {
    printf("Please write %d numbers: \n", n);
  }

  // i asked for numbers from user.
  for (int i = 0; i < n; ++i) {
    scanf(" %d", &number[i]);
  }

  // I asked for a number to search from the inputted array.
  printf("Please enter numbers to search: \n");
  scanf("%d", &a);

  n = n - 1;
  for (int a = 0; a < n; ++a) {
    scanf("%d", &ara[a]);
  }

  n = n + 1;
  for (i = 0; i < n; ++i) {  // my program looks every line to find the number.

    for (a = 0; a < n; ++a) {
      // if program can find a number equals to the ones in array, it will print
      // the line.
      if (number[i] == ara[a]) {
        // i wrote i+1 because normally number 1 is considered as number 0 in
        // arrays.
        printf("%d is at position %d in the array.\n", ara[a], i + 1);
        // break; //to stop the loop.
      }
    }
  }
}

#包括
#包括
int main（）{
int i，b，c，n，a；
国际原子能机构[200]；
整数[200]；
printf（“请输入您要输入的数字：\n”）；
scanf（“%d”和“&n”）；
如果（n>200）{
//我限制了用户可以输入的数量。
printf（“你不能写超过200，超出范围。”）；
}否则{
printf（“请填写%d个数字：\n”，n）；
}
//我向用户询问电话号码。
对于（int i=0；i

您的程序未定义为读取未初始化的内存。我可以证明，无论这个程序的输入是什么，它最终都会读取

ara[n-1]

，由于行

n=n-1，它从未初始化过和
n=n+1。如果编译器成功地完成了这个归纳，它可以删除整个嵌套的
for
循环，因为现代编译器假定未定义的行为意味着无法访问的代码

此外，此扫描行错误：
scanf（“%d”和&a）读取
a
不起作用。读入
n
也是如此。您需要另一个整数来存储搜索数组中的数字数量。你不能同时使用
n
。
我已经在你的代码上修改了几行，现在它可以正常工作了请自己检查：

#include <stdio.h>
#include <stdlib.h>
int main() {
  int i, b, c, n, a;
  int ara[200];
  int number[200];

  printf("Please enter how many numbers you wanna input: \n");
  scanf("%d", &n);
  if (n > 200) {
    // i limited the numbers users can input.
    printf("You can't write higher than 200. It's out of range. ");
  } else {
    printf("Please write %d numbers: \n", n);
  }

  // i asked for numbers from user.
  for (int i = 0; i < n; ++i) {
    scanf(" %d", &number[i]);
  }

  // I asked for a number to search from the inputted array.
  printf("Please enter numbers to search: \n");
//   scanf("%d", &a); // Your major problem was beginning from here.

  for (int a = 0; a < n; ++a) {
    scanf("%d", &ara[a]);
  }

  for (i = 0; i < n; ++i) {  // my program looks every line to find the number.

    for (a = 0; a < n; ++a) {
      // if program can find a number equals to the ones in array, it will print
      // the line.
      if (number[i] == ara[a]) {
        // i wrote i+1 because normally number 1 is considered as number 0 in
        // arrays.
        printf("%d is at position %d in the array.\n", ara[a], i + 1);
        // break; //to stop the loop.
      }
    }
  }
  
}

更新以提高可读性和效率

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

int main() {
  int i, b, c, n, a;
  int ara[200];
  int number[200];

  printf("Please enter how many numbers you wanna input: ");
  scanf("%d", &n);
  if (n > 200) {
    // i limited the numbers users can input.
    printf("You can't write higher than 200. It's out of range. ");
  } else {
    printf("Please write %d numbers:\n", n);
  }

  // i asked for numbers from user.
  for (int i = 0; i < n; ++i) {
    printf("%d. number: ",i+1);
    scanf(" %d", &number[i]);
  }

  // First ask how many numbers the user wants to search.
  printf("\nHow many numbers will be searched?: ");
  scanf("%d", &a);

  for (int s = 0; s < a; ++s) {
    printf("%d. number to search: ",s+1);
    scanf("%d", &ara[s]);
  }

  /* A better approach is to iterate in the outer loop for each number to be searched
   * instead of iterateing for the numbers in the list.
   */
  bool found;
  for (i = 0; i < a; ++i) {  // In the outer loop iterate for the numbers to be searched.
    found = false; // Set found to false for each new number to be searched 
    for (b = 0; b < n; ++b) { // In the inner loop iterate for the list
      // if program can find a number equals to the ones in array, it will print
      // the line.
      if (number[b] == ara[i]) {
        // i wrote i+1 because normally number 1 is considered as number 0 in
        // arrays.
        printf("%d is at position %d in the array.\n", ara[i], b + 1);
        found = true; // set found to true if you find it
      }
    }
    
    if(!found){
        printf("The number <%d> does not exist in the list\n",ara[i]);
    }
  }
  
} 

#包括
#包括
#包括
int main（）{
int i，b，c，n，a；
国际原子能机构[200]；
整数[200]；
printf（“请输入您要输入的数字：”）；
scanf（“%d”和“&n”）；
如果（n>200）{
//我限制了用户可以输入的数量。
printf（“你不能写超过200，超出范围。”）；
}否则{
printf（“请填写%d个数字：\n”，n）；
}
//我向用户询问电话号码。
对于（int i=0；i

更新的代码输出如下：

[kozmotronik@kozmotronik-ge600nc0nd Belgeler]$ ./karalama
Please enter how many numbers you wanna input: 
4
Please write 4 numbers: 
3
5
3
7
Please enter numbers to search: 
3
4
5
7
3 is at position 1 in the array.
5 is at position 2 in the array.
3 is at position 3 in the array.
7 is at position 4 in the array.

Please enter how many numbers you wanna input: 8
Please write 8 numbers:
1. number: 100
2. number: 2345
3. number: 457
4. number: 234
5. number: 2
6. number: 657
7. number: 4
8. number: 900

How many numbers will be searched?: 4
1. number to search: 4
2. number to search: 23
3. number to search: 657
4. number to search: 15
4 is at position 7 in the array.
The number <23> does not exist in the list
657 is at position 6 in the array.
The number <15> does not exist in the list

请输入您要输入的数字：8
请写8个数字：
1.电话：100
2.电话：2345
3.电话：457
4.电话：234
5.编号：2
6.电话：657
7.电话：4
8.电话：900
将搜索多少个数字？：4
1.搜索号码：4
2.搜索号码：23
3.搜索号码：657
4.搜索号码：15
4位于阵列中的位置7。
列表中不存在该编号
657位于阵列中的位置6。
列表中不存在该编号
非常感谢你，我是个新手，所以我迷路了，我正在努力学习。正如你所说，我重新安排了“数字搜索部分”，并放置了新的整数来存储我的值，现在它开始工作了。这真的很有帮助！非常感谢你！作为初学者，我很难看出自己的错误。这么有帮助的回答！您的代码中有一个更新。过来看！