Question

我收到了制作字典的任务，该字典最多可包含2个每个单词的定义，该定义必须能够包含空格。

我已经搜索并查看了许多使用获取，fgets，修改"%s"运算符以包含多个ASCII值的示例。

但问题是，现在它们对我有用，当程序应该获得用户输入时，它似乎只是跳过这一行 - 我在Code Blocks和VS 2012中都尝试过。这是代码的相关部分，请注意字符串大小temp_word [81], temp_definition [201]以及words, definitions为char**存在限制。

printf("Please enter word no.%d and how many difinitions will it have(1/2):\n", i+1);
scanf("%s %d", temp_word, &no_definitions); //gets word and no. of difinitons
words [i] = (char*) malloc (strlen(temp_word)+1); //memory is allocated according to length of the word, + 1 for \0 that is'n counted by the function
strcpy(words [i] , temp_word);

printf("\n\n%d, %s\n", no_definitions, temp_word); //debugging
printf("Please enter definition no.1:\n");
scanf("%99[^\n]s", temp_definitoin);
definitions_a [i] = (char*) malloc (strlen(temp_definitoin)+1); //memory is allocated according to length of the definition, + 1 for \0 that is'n counted by the function

if (definitions_a [i] == NULL) //must take malloc's failure into account (NULL!)
{
    printf ("Memory allocation failed!\n");
    goto exit;
}

strcpy(definitions_a [i], temp_definitoin);

if (no_definitions == 1)
{
    definitions_b [i] == NULL; //if only one definition is entered the second one will hold the NULL pointer
}
else if (no_definitions == 2)
{
    printf("Please enter definition no.2:\n");
    gets(temp_definitoin); //gets definition
    definitions_b [i] = (char*) malloc (strlen(temp_definitoin)+1); //memory is allocated according to length of the definition, + 1 for \0 that is'n counted by the function

    if (definitions_a [i] == NULL) //must take malloc's failure into account (NULL!)
    {
        printf ("Memory allocation failed!\n");
        goto exit;
    }

    strcpy(definitions_b [i] , temp_definitoin);
}

//}

Answer 1

BTW scanf("%99[^\n]s", temp_definitoin); - ＆gt; scanf(" %99[^\n]", temp_definitoin);（添加空格，没有s） - chux

从用户那里获取带有空格的字符串

1 个答案: