如何将字符从变量放入C中的数组？

Question

我想将一个字符从一个变量放到C中的一个字符数组中。另外我想打印后面的反转数组，因为您可以看到但现在不是问题。

这是迄今为止我已经得到了代码：

作为标准输入我使用的是一个txt文件与“< input.txt中”作为命令行参数，它有57个字符。

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

int main() 
{ 
    int counter = 0; 
    char character_array[57]; 
    int i = 0; 
    int j = 0; 
    char character = 0; 

    // While EOF is not encountered read each character 
    while (counter != EOF) 
    { 
    // Print each character 
    printf("%c", counter); 
    // Continue getting characters from the stdin/input file 
    counter = getchar(stdin); 
    // Put each character into an array 
    character_array[j] = { counter }; 
    j = j + 1; 
    } 

    // Print the array elements in reverse order 
    for (i = 58; i > 0; i--) 
    { 
    character = character_array[i]; 
    printf("%c", character); 
    } 

    return 0; 
} 

我的IDE在第一个花括号“预期表达式”之后的第35行说。

// Put each character into an array 
    character_array[j] = { counter }; 

所以我想它失败了。我假设我不能像数组中那样放置字符变量？否则我会怎么做呢？

PS：我是新来的C.

Answer 1

在该行取出{和}，使它看起来像：

character_array[j] = counter ; 

改进的代码：

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

int main() 
{ 
    int counter = 0; 
    char character_array[57]; 
    int i = 0; 
    int j = 0; 
    //char character = 0; Unused variable 

    // While EOF is not encountered read each character 
    while ((counter = getchar()) != EOF) 
    { 
    // Print each character 
    printf("%c", counter); 
    character_array[j] = counter; 
    j++; 
    } 
    for (i = j - 1; i >= 0; i--) /* Array indices start from zero, and end at length - 1 */ 
    { 
    printf("%c", character_array[i]); 
    } 
} 

Answer 2

character_array[j] = counter; 

就这么简单，我认为