c - 在C中: How to print function argument

Question

我想在 printf 命令中打印函数参数。请大家帮忙推荐一下。

void printline(char ch, int len);
value(float, float, int);

main()
{
    double amount;
    printline('=', 30);
    amount = value(500, 0.12, 5); //  I want to print argument of function value. please help
    printf("The total amount is: %f \n", amount);
    //printf("%f\t%f\t%d\t%f \n", 500, 0.12, 5, amount);
    printline('=', 30);
    _getch();
}

void printline(char ch, int len)
{
    int i;
    for (i = 0; i < len; i++)
        printf("%c", ch);
    printf("\n");
}

value(float p, float r, int n)
{
    int year;
    float sum;
    sum = p;
    year = 1;
    while (year <= 5)
    {
        sum = sum * (1 + r);
        year = year + 1;
    }
    return(sum);
}

Answer 1

在您的 printline 函数中，您只有一个字符作为参数。所以迭代它是没有意义的。如果要打印字符串或字符数组，则需要使用char *或char[]并对其进行迭代。所以你的函数 printline 可能如下所示:

void printline(char *ch, int len)
{
    int i;
    for (i = 0; i<len; i++)
        printf("%c", ch[i]);
    printf("\n");
}

只需确保 len 不大于 *ch 的长度即可。

更好的解决方案是，您不必担心 len 的值，而是逐个打印字符，直到遇到 \0 字符表示数组的结尾。

 void printline(char *ch)
    {
        int i;
        for (i = 0; ch[i] != '\0'; ++i)
            printf("%c", ch[i]);
        printf("\n");
    }

或者even better

If the string is null terminated, use printf("%s", ch). If the string is not null terminated but the length is supplied, use printf("%.*s", len, ch). In both cases, there's no loop in the user code; the loop is buried inside the printf() function. Further, since there's a newline printed after the loops, use printf("%s\n", ch) or printf("%.*s\n", len, ch) and skip the extra printf() after the loop.