strtok_s 忽略第一个字符

Question

我正在尝试获取一个输入并将其转换为四个变量名称。

我在每 strtok_s 之后进行检查以查看我得到了什么，但第一个单词只在 4 个字符之后才算。

我的代码：

void zeros()
{
    char buffer[81];
    fgets(buffer, sizeof(buffer), stdin);
    printf("%s\n", buffer);
    char *command = strtok_s(buffer, " \t", &buffer);
    printf_s("the command you selcted is %s\n", command);
    char *Matname = strtok_s(NULL, " \t", &buffer);
    printf_s("the name you selcted is %s\n", Matname);
    char *row = strtok_s(NULL, "  \t", &buffer);
    printf_s("the rowsize you selcted is %s\n", row);
    char *col = strtok_s(NULL, " \t", &buffer);
    printf_s("the colsize you selcted is %s\n", col);
    return 0;
}

Answer 1

strtok_s的最后一个参数不正确，应该是strtok_s用来存储其内部状态的指针。

您的函数应该更像这样：

void zeros() // or int zeros() if it's supposed to return int
{
    char buffer[81];
    char *ptr;
    fgets(buffer, sizeof buffer, stdin);
    printf("%s\n", buffer);
    char *command = strtok_s(buffer, " \t", &ptr);
    printf_s("the command you selcted is %s\n", command);
    char *Matname = strtok_s(NULL, " \t", &ptr);
    printf_s("the name you selcted is %s\n", Matname);
    char *row = strtok_s(NULL, "  \t", &ptr);
    printf_s("the rowsize you selcted is %s\n", row);
    char *col = strtok_s(NULL, " \t", &ptr);
    printf_s("the colsize you selcted is %s\n", col);
    // return 0; // if function return type is int
}

另一个问题，尽管不是主要问题，是该函数具有 void return 类型和 return 一个 int.

strtok_s 忽略第一个字符

strtok_s ignors first characters

c

char

strtok