c - 删除C中字符串中出现的每个\n

Question

我已经看过这方面的教程和提示，但它们通常不适用于 C。这是我到目前为止所写的内容:

    printf("\nBefore removing: %s", an_name);
    printf("%d", strlen(an_name));

    for (i=0;i<strlen(an_name);i++){
        printf("\nthe loop: %d",i);

        if (an_name[i]!='\n'){
        new_name[i] = an_name[i];
        printf("\nthe loop: %d", i);
    }
    else{//when a \n is detected
        new_name[i] = an_name[i+1];
        i=i+1;
    }   
    printf("\nAfter removing: %s\n", new_name);
    printf("%d", strlen(an_name));
    getchar();

    return 0;
}

It does half the job. I want to remove the \n character and put all the new characters in a new string. But how?

Answer 1

您需要输入数组和输出数组的单独索引。否则，您永远不会将任何内容放入输出数组中跳过的换行符所在的位置，因此那里有一个未初始化的值。在您的测试中，它恰好包含 0，这是字符串终止符，因此它只打印了之前的所有内容。

循环完成后，您还需要在输出字符串的末尾添加一个空字节。

您可以使用 isspace() 来测试是否有空格。

#include <ctype.h>
#include <string.h>
#include <stdio.h>

int main(){
    char an_name[20], new_name[20];
    int i;

    strcpy(an_name,"Ant\nho ny");

    printf("\nBefore removing: %s\n", an_name);
    printf("%ld", strlen(an_name));

    int j = 0;
    for (i=0;i<strlen(an_name);i++){
        printf("\nthe loop: %d",i);

        if(!isspace(an_name[i])){
            new_name[j++] = an_name[i];
        }
    }
    new_name[j] = '\0';

    printf("\nAfter removing: %s\n", new_name);
    printf("%ld", strlen(an_name));
    getchar();

    return 0;
}

输出:

Before removing: Ant
ho ny
9
the loop: 0
the loop: 1
the loop: 2
the loop: 3
the loop: 4
the loop: 5
the loop: 6
the loop: 7
the loop: 8
After removing: Anthony
9