一般来说，与其删除不想要的数据，不如从一开始就避免写入数据。如果你不想在缓冲区中使用换行符，不要使用fgets。相反，请使用getc或fgetc或scanf。也许是这样的:

#include <stdio.h>
#include <stdlib.h>
int
main(void)
{
        char Name[256];
        char fmt[32];
        if( snprintf(fmt, sizeof fmt, "%%%zd[^\n]", sizeof Name - 1) >= (int)sizeof fmt ){
                fprintf(stderr, "Unable to write format\n");
                return EXIT_FAILURE;
        }
        if( scanf(fmt, Name) == 1 ) {
                printf("Name = %s\n", Name);
        }
        return 0;
}

注意，这种特殊的方法将不读取换行符，因此您可能希望使用像“%255[^\n]%*c”这样的格式字符串来丢弃它(例如，sprintf(fmt，“%%%zd[^\n]%%*c”，sizeof Name - 1);)，或者可能在scanf后面加上getchar()。

下面是一个从fgets()保存的字符串中删除潜在'\n'的快速方法。 它使用strlen()，带有2个测试。

char buffer[100];
if (fgets(buffer, sizeof buffer, stdin) != NULL) {

  size_t len = strlen(buffer);
  if (len > 0 && buffer[len-1] == '\n') {
    buffer[--len] = '\0';
  }

现在根据需要使用buffer和len。

该方法的附带好处是为后续代码提供len值。它可以比strchr(Name， '\n')更快。参考YMMV，但两种方法都有效。

在某些情况下，原始fgets()将不包含在"\n"中: A)行太长，不适合buffer，所以只有'\n'前面的char被保存在buffer中。未读字符保留在流中。 B)文件的最后一行没有以“\n”结尾。

如果输入在某个地方嵌入了空字符'\0'，strlen()报告的长度将不包括'\n'位置。

其他一些回答的问题:

strtok(buffer, "\n"); fails to remove the '\n' when buffer is "\n". From this answer - amended after this answer to warn of this limitation. The following fails on rare occasions when the first char read by fgets() is '\0'. This happens when input begins with an embedded '\0'. Then buffer[len -1] becomes buffer[SIZE_MAX] accessing memory certainly outside the legitimate range of buffer. Something a hacker may try or found in foolishly reading UTF16 text files. This was the state of an answer when this answer was written. Later a non-OP edited it to include code like this answer's check for "". size_t len = strlen(buffer); if (buffer[len - 1] == '\n') { // FAILS when len == 0 buffer[len -1] = '\0'; } sprintf(buffer,"%s",buffer); is undefined behavior: Ref. Further, it does not save any leading, separating or trailing whitespace. Now deleted. [Edit due to good later answer] There are no problems with the 1 liner buffer[strcspn(buffer, "\n")] = 0; other than performance as compared to the strlen() approach. Performance in trimming is usually not an issue given code is doing I/O - a black hole of CPU time. Should following code need the string's length or is highly performance conscious, use this strlen() approach. Else the strcspn() is a fine alternative.

也许最简单的解决方案是使用我最喜欢的一个鲜为人知的函数strcspn():

buffer[strcspn(buffer, "\n")] = 0;

如果你想让它也处理'\r'(比如，如果流是二进制的):

buffer[strcspn(buffer, "\r\n")] = 0; // works for LF, CR, CRLF, LFCR, ...

该函数会计算字符的数量，直到遇到'\r'或'\n'(换句话说，它会找到第一个'\r'或'\n')。如果它没有碰到任何东西，它就会停在'\0'(返回字符串的长度)。

请注意，即使没有换行符，这也可以正常工作，因为strcspn在'\0'处停止。在这种情况下，整行代码只是将“\0”替换为“\0”。

对于单个“\n”切边，

void remove_new_line(char* string)
{
    size_t length = strlen(string);
    if((length > 0) && (string[length-1] == '\n'))
    {
        string[length-1] ='\0';
    }
}

对于多个“\n”切边，

void remove_multi_new_line(char* string)
{
  size_t length = strlen(string);
  while((length>0) && (string[length-1] == '\n'))
  {
      --length;
      string[length] ='\0';
  }
}

 for(int i = 0; i < strlen(Name); i++ )
{
    if(Name[i] == '\n') Name[i] = '\0';
}

你应该试试。这段代码基本上遍历字符串，直到找到'\n'。当它被找到时，'\n'将被空字符结束符'\0'取代

注意，在这一行中比较的是字符而不是字符串，那么就不需要使用strcmp():

if(Name[i] == '\n') Name[i] = '\0';

因为你将使用单引号而不是双引号。如果你想了解更多，这里有一个关于单引号和双引号的链接

我的新手方式;-)请让我知道这是正确的。这似乎对我所有的案子都有效:

#define IPT_SIZE 5

int findNULL(char* arr)
{
    for (int i = 0; i < strlen(arr); i++)
    {
        if (*(arr+i) == '\n')
        {
            return i;
        }
    }
    return 0;
}

int main()
{
    char *input = malloc(IPT_SIZE + 1 * sizeof(char)), buff;
    int counter = 0;

    //prompt user for the input:
    printf("input string no longer than %i characters: ", IPT_SIZE);
    do
    {
        fgets(input, 1000, stdin);
        *(input + findNULL(input)) = '\0';
        if (strlen(input) > IPT_SIZE)
        {
            printf("error! the given string is too large. try again...\n");
            counter++;
        }
        //if the counter exceeds 3, exit the program (custom function):
        errorMsgExit(counter, 3); 
    }
    while (strlen(input) > IPT_SIZE);

//rest of the program follows

free(input)
return 0;
}