c - 在 C 中读取并打印字符串

Question

所以我一直在尝试读取输入字符串，然后通过不使用 scanf() 来打印它也不printf()但是getchar()和 putchar()反而。程序似乎陷入了循环，我无法发现错误。

#include <stdio.h>

void getstring(char *c)
{
char inS[100];
char n;
int i = 0;  
while ((n = getchar()) != '\n') {
    inS[i] = n;
    i++;        
}
inS[i] = '\0';
i = 0;
while (inS[i] != '\0') {
    putchar(inS[i]);
    i++;
}
}
main()
{
char *prompt;
prompt = "Enter a sentence: \n";
getstring(&prompt);
printf("%s", prompt);

}

Answer 1

您的代码有一些问题。这是更正后的代码:

#include <stdio.h>

void getstring(void) /* You don't use the passed argument. So, I've removed it */
{
    char inS[100];
    /* `char n;` getchar() returns an int, not a char */
    int n;
    int i = 0;  
    while (i < 99 && (n = getchar()) != '\n' && n != EOF)  /* Added condition for preventing a buffer overrun and also for EOF */
    {
        inS[i] = n;
        i++;        
    }
    inS[i] = '\0';
    putchar('\n'); /* For seperating input and output in the console */
    i = 0;
    while (inS[i] != '\0')
    {
        putchar(inS[i]);
        i++;
    }
    putchar('\n'); /* For cleanliness and flushing of stdout */
}
int main(void) /* Standard signature of main */
{
    char *prompt;
    prompt = "Enter a sentence: \n";
    printf("%s", prompt); /* Interchanged these two lines */
    getstring(); /* Nothing to pass to the function */
    return 0; /* End main with a return code of 0 */
}

注意:输入循环将在以下任一情况下结束

• 遇到\n(Enter)。
• 遇到了EOF。
• 从 stdin 中读取了 99 个字符。