Question

嗨我知道有很多例子可以输入一个字符并输出字符的位置。但是，你如何在输入位置并输出角色的其他方面做到这一点？

输入字符串：abcdefghijklmnopqrstuvwxyz
输入位置：0
输出：位置0的字符是'a'

我使用atoi吗？

int main(void) {

char line[SIZE];
int position;

// int character = getchar(); //get string
scanf("%d\n", &position); //get position
int i = 0;
while(fgets(line, SIZE, stdin) != NULL) {
    if(line[i] == position) { 
       printf("The character in postion %d is '%c'\n", position,line[i]);   
        //flag = TRUE;
   // }
    i++;
}       

return 0;
}

Answer 1

while(fgets(line, SIZE, stdin) != NULL) 

  {

     line[strlen(line)-1] = '\0'; // as fgets append '\n' before '\0', replacing '\n' with '\0'
     if(strlen(line) > position)
     {   
           printf("The character in postion %d is '%c'\n", position,line[position]); 
     }   
     else
     {   
        printf("invalid position\n");
     }   

}

Answer 2

你可能想要这个：

#include <stdio.h>

#define SIZE 100

int main(void) {    
  char line[SIZE];
  int position;

  scanf("%d", &position); //get position
  getchar();   // absorb \n from scanf (yes scanf is somewhat odd)

  while (fgets(line, SIZE, stdin) != NULL) {
      printf("The character in postion %d is '%c'\n", position, line[position]);
  }

  return 0;
}

为了简洁起见，没有超出范围的检查

执行示例：

1
abc
The character in postion 1 is 'b'
Hello
The character in postion 1 is 'e'
TEST
The character in postion 1 is 'E'

这个小例子也有帮助：

#include <stdio.h>

#define SIZE 100

int main(void) {
  char line[SIZE];

  fgets(line, SIZE, stdin);
  for (int i = 0; line[i] != 0; i++)
  {
    printf("The character in postion %d is '%c'\n", i, line[i]);
  }
}

为简洁起见，没有进行错误检查。

通过在c中输入位置来查找字符串的字符

2 个答案: