在getchar()之后立即使用fgets()

我可以通过添加一个附加的fgets()(由于某种原因而必须将一个大于1个char的字符串传递给它)来解决此问题，但是有解决此问题的更优雅的方法吗?解决方案好，您可以使用 scanf(％* [\ n]"); 忽略任何数字连续的换行符.或 scanf(％* 1 [\ n]"); 只吃一个 换行符.如果第一个字符是其他任何字符，则不会使用.另一种选择是使用低级操作 getchar 和 ungetc : int eat_stdin_newline(void){int ch = getchar();if(ch！= EOF&&ch！='\ n'){//如果不是EOF或换行符，请将其推回...ungetc(ch，stdin);//必须成功}返回ch} 然后您可以在任何需要的地方调用此函数: eat_stdin_newline(); voidread_stdin(trace_t* trace, state_t state, action_t** action_list) { // initial stage int c; while ((c = getchar())!= EOF && c!='#') { if (my_isalpha(c)==LOWERCASE) { state[c-ASCII_CODE_LOWER_A] = '1'; } } printf("%s\n", state); char str[2]; fgets(str, 2, stdin); printf("%s", str);}If '#' is the last character I enter in the getchar() loop, fgets() records the newline character from when I press enter and skips to the print statement immediately (which prints the '\n')I could fix this by adding an additional fgets()(which has to have a string that is longer than 1 char passed to it for some reason?) but is there a more elegant way of solving this? 解决方案 Well, you can use scanf("%*[\n]"); to ignore any number of consecutive newline. Or scanf("%*1[\n]"); for eating only one newline. If any other character is the first one, it is not consumed.Another option would be to use low-level operations getchar and ungetc:int eat_stdin_newline(void) { int ch = getchar(); if (ch != EOF && ch != '\n') { // if it wasn't EOF or newline, push it back... ungetc(ch, stdin); // must succeed } return ch;}Then you can call this function wherever you want:eat_stdin_newline(); 这篇关于在getchar()之后立即使用fgets()的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！