C语言处理多行字符串输入遇问题：代码无输出请求排查

Let's break down the problem with your code first, then walk through solutions to get it working as expected.

The Root Cause #

When you use scanf("%d", &t) to read the integer 3, it leaves the newline character (\n) you pressed after entering the number in the input buffer.

Your next line uses scanf("%[^\n]%*c", a) to read a line of text. Unlike other format specifiers like %s, the %[^\n] pattern does not skip leading whitespace. So it immediately hits the leftover newline character, fails to match any text, and returns 0 instead of 1. This means your loop never runs—hence, no output.

Fixes to Try #

Fix 1: Consume the Leftover Newline #

Add a getchar() right after reading the integer to eat the newline left in the buffer:

#include <stdio.h>
int main(){
    int t;
    scanf("%d",&t);
    getchar(); // Reads the newline after the integer
    char a[100];
    while(scanf("%[^\n]%*c",a) == 1){
        printf("%s\n",a);
        --t;
        if(t == 0) break;
    }
    return 0;
}

Fix 2: Skip Whitespace in the String Scan #

Modify the format string to include a leading space—this tells scanf to skip any leading whitespace (including newlines and spaces) before reading the string:

#include <stdio.h>
int main(){
    int t;
    scanf("%d",&t);
    char a[100];
    // Leading space skips whitespace before reading the line
    while(scanf(" %[^\n]%*c",a) == 1){
        printf("%s\n",a);
        --t;
        if(t == 0) break;
    }
    return 0;
}

Fix 3: Use fgets for Safer Line Reading #

scanf("%[^\n]") can cause buffer overflow if the line is longer than your array size. fgets is safer because it enforces a character limit. Here's how to use it:

#include <stdio.h>
#include <string.h> // For strcspn

int main(){
    int t;
    scanf("%d",&t);
    getchar(); // Consume the newline after the integer
    char a[100];
    while(fgets(a, sizeof(a), stdin) != NULL && t > 0){
        // Remove the trailing newline that fgets includes
        a[strcspn(a, "\n")] = '\0';
        printf("%s\n",a);
        --t;
    }
    return 0;
}

For Your Specific Example Input #

If you only want to print the single line after the integer (ignoring the loop for t), simplify the code like this:

#include <stdio.h>
int main(){
    int t;
    scanf("%d", &t);
    char a[100];
    scanf(" %[^\n]", a); // Skips space after 3, reads rest of the line
    printf("%s\n", a);
    return 0;
}

This will output exactly the string you expect from your example input.

内容的提问来源于stack exchange，提问作者chelsea