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.

Tim Čas one line对于调用fgets获得的字符串来说非常神奇，因为您知道它们在末尾包含一个换行符。

如果您处于不同的上下文中，并且希望处理可能包含多个换行符的字符串，则可能需要查找strrspn。它不是POSIX，这意味着您不会在所有unix上找到它。我为自己的需要写了一个。

/* Returns the length of the segment leading to the last 
   characters of s in accept. */
size_t strrspn (const char *s, const char *accept)
{
  const char *ch;
  size_t len = strlen(s);

more: 
  if (len > 0) {
    for (ch = accept ; *ch != 0 ; ch++) {
      if (s[len - 1] == *ch) {
        len--;
        goto more;
      }
    }
  }
  return len;
}

对于那些在C中寻找等效Perl chomp的人来说，我认为这就是它(chomp只删除后面的换行符)。

line[strrspn(string, "\r\n")] = 0;

strrcspn函数:

/* Returns the length of the segment leading to the last 
   character of reject in s. */
size_t strrcspn (const char *s, const char *reject)
{
  const char *ch;
  size_t len = strlen(s);
  size_t origlen = len;

  while (len > 0) {
    for (ch = reject ; *ch != 0 ; ch++) {
      if (s[len - 1] == *ch) {
        return len;
      }
    }
    len--;
  }
  return origlen;
}

这就是我的解。非常简单。

// Delete new line
// char preDelete[256]  include "\n" as newline after fgets

char deletedWords[256];
int iLeng = strlen(preDelete);
int iFinal = 0;
for (int i = 0; i < iLeng; i++) {
    if (preDelete[i] == '\n') {

    }
    else {
        deletedWords[iFinal]  = preDelete[i];
        iFinal++;
    }
    if (i == iLeng -1 ) {
        deletedWords[iFinal] = '\0';
    }
}

如果可以选择使用getline(不要忽略其安全性问题)，如果希望使用大括号指针，则可以避免使用字符串函数，因为getline返回的是字符数。如下图所示

#include <stdio.h>
#include <stdlib.h>
int main()
{
    char *fname, *lname;
    size_t size = 32, nchar; // Max size of strings and number of characters read
    fname = malloc(size * sizeof *fname);
    lname = malloc(size * sizeof *lname);
    if (NULL == fname || NULL == lname)
    {
        printf("Error in memory allocation.");
        exit(1);
    }
    printf("Enter first name ");
    nchar = getline(&fname, &size, stdin);
    if (nchar == -1) // getline return -1 on failure to read a line.
    {
        printf("Line couldn't be read..");
        // This if block could be repeated for next getline too
        exit(1);
    }
    printf("Number of characters read :%zu\n", nchar);
    fname[nchar - 1] = '\0';
    printf("Enter last name ");
    nchar = getline(&lname, &size, stdin);
    printf("Number of characters read :%zu\n", nchar);
    lname[nchar - 1] = '\0';
    printf("Name entered %s %s\n", fname, lname);
    return 0;
}

注意:getline的[安全问题]不应该被忽视。

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”。

如果每一行都有'\n'，则直接从fgets输出中删除'\n'

line[strlen(line) - 1] = '\0';

否则:

void remove_newline_ch(char *line)
{
    int new_line = strlen(line) -1;
    if (line[new_line] == '\n')
        line[new_line] = '\0';
}