Question

我正在编写一个基本程序来使用C命令行参数来计算算术几何平均值。但是程序似乎没有识别我输入的任何内容。这是我的代码：

/*
*This program will calculate an arithmetic-geometric mean
*provided two numbers (x,y) and an epsilon (e) are entered.
*/

#include<stdlib.h>
#include<stdio.h>
#include<math.h>

int main (int argc, char **argv) {

    //Check for command line argument.
    if (argc != 4) {
        printf ("Please enter two numbers (x,y) and an epsilon (e)\n");
        printf ("as command line arguments for an AGM calculation\n");
        exit(1);
    }

    double e,x,y,an,gn;

    x = atof (argv[1]); //First number x.
    y = atof (argv[2]); //Second number y.
    e = atof (argv[3]); //Number of repetitions e.

    double absoluteAnMinusGn; //Continuation condition.
    double a = (x + y) / 2; //Normal arithmetic mean.
    double g = sqrt (x * y); //Normal geometric mean.

    an = (a + g) / 2; //Iteration 1 for calculation arithmetic mean.
    gn = sqrt (a * g); //Iteration 1 for calculation geometric mean.
    absoluteAnMinusGn = an - gn; //Calculates continuation condition.
       if (absoluteAnMinusGn < 0) {
           absoluteAnMinusGn = absoluteAnMinusGn * (-1); //Ensures absolute value of continuation condition.
        }

    printf ("DEBUG IN: x%d, y%d, e%d, absoulteAnMinusGn%d, a%d, g%d, an%d, gn%d\n", x,y,e,absoluteAnMinusGn,a,g,an,gn);//DEBUG CODE 

    while (absoluteAnMinusGn > e) {
        an = (a + g) / 2;
        gn = sqrt (a * g);
        a = an;
        g = gn;
        absoluteAnMinusGn = an - gn;
        if (absoluteAnMinusGn < 0) {
            absoluteAnMinusGn = absoluteAnMinusGn * (-1);
        }
    }

    //printf ("The arithmetric-geometric mean is (%d,%d) for %d\n", a,g,e); 
            printf ("DEBUG OUT: x%d, y%d, e%d, absoulteAnMinusGn%d, a%d, g%d, an%d, gn%d\n", x,y,e,absoluteAnMinusGn,a,g,an,gn);//DEBUG CODE 

    return 0;
}

我在命令行中输入以下内容：agm.exe 3 4 5

我得到以下输出：

DEBUG IN: x0, y10742661112, e0, absoluteANMinusGn1074790400, a0, g1075052544, an-171951648, gn1057505593

DEBUG OUT: x0, y10742661112, e0, absoluteANMinusGn1074790400, a0, g1075052544, an-171951648, gn1057505593

我昨天制作了一个类似的程序，用于使用命令行输入计算积分，这些输入完全符合预期。代码就在这里：

/*
*This program will calculate a Riemann sum using the 
*left hand rule for sin(x)/x.
*/

#include<stdlib.h>
#include<stdio.h>
#include<math.h>

int main (int argc, char **argv) {

    //Check for command line argument.
    if (argc != 4) {
        printf ("Please enter integral bounds (a,b) and number of intervals (n)\n");
        printf ("as command line arguments for a Riemann sum calculation \n");
        exit(1);
    }

    double a,b,i,n,h,riemann,rectangles,answer;

    a = atof (argv[1]); //Lower bound of integral.
    b = atof (argv[2]); //Upper bound of integral.
    n = atof (argv[3]); //Number of intervals.

    h = (b - a) / n; //Delta X.
    i = 0; //Counts intervals.

    //Calculation of Left Hand Riemann Sum.
    while (i <= (n - 1)) {
        if (a == 0 && i == 0) { //Stops from dividing by zero.
            rectangles = 1;
            i++;
        }
        riemann = (sin(a + (i * h))) / (a + (i * h));
        rectangles += riemann;
        i++;
    }
    //Finalize answer.
    answer = rectangles * h;

    printf ("Sin(x)/x for bounds (%f , %f) with %f intervals is approximately %f \n", a,b,n,answer);

    return 0;
}

左手Riemann sum的上述程序正确输出，几乎与我对AGM的代码相同。有人可以帮我弄清楚这里出了什么问题吗？我到处搜索过，找不到解决办法。我知道AGM代码可能设置为输出错误答案，但我主要担心的是修复命令行参数识别。我可以稍后重做我的数学。

Answer 1

打印double的格式说明符是%f。如果你没有提供正确的格式说明符来处理double，并且将double传递给%d格式说明符 - 它会导致未定义的行为。（%d期望积分参数不是double）（在你的中）案例错误输出）。

来自§7.21.6.3p9 N1570（c11标准）

如果任何参数不是相应转换规范的正确类型，则行为未定义。

Answer 2

只需用List<User>（这是double的格式说明符）替换<{strong> List<User> temp = new ArrayList();（这是整数的格式说明符），因为您使用的是类型变量加倍，不是整数。

Answer 3

我没有测试你的算法，但我在你的代码中发现了至少三个问题。我现在列出如下：

忘记初始化你的变量，这就是你得到Large Numbers内存未定义数据的原因！
没有将你的argv转换为整数类型。
printf的格式错误。

我已将我的固定代码放在此处，您可以对其进行测试。如果它可以像你期望的那样运行！

#include<stdlib.h>
#include<stdio.h>
#include<math.h>

int main (int argc, char *argv[]) {
    if (argc != 4) {
        printf ("Please enter two numbers (x,y) and an epsilon (e)\n");
        printf ("as command line arguments for an AGM calculation\n");
        exit(1);
    }

    //  argv is pointer which pointer to pointer 
    // and argv[1] is a pointer to a string.
    // so we need to covert string to int
    char *p=NULL;
    int x = strtol(argv[1], &p, 10);
    int y = strtol(argv[2], &p, 10);
    int e = strtol(argv[3], &p, 10);

    //Check for command line argument.

    double an = 0.0,gn =0.0;


    double absoluteAnMinusGn=0.0; //Continuation condition.
    double a = (x + y) / 2; //Normal arithmetic mean.
    double g = sqrt (x * y); //Normal geometric mean.

    an = (a + g) / 2; //Iteration 1 for calculation arithmetic mean.
    gn = sqrt (a * g); //Iteration 1 for calculation geometric mean.
    absoluteAnMinusGn = an - gn; //Calculates continuation condition.
       if (absoluteAnMinusGn < 0) {
           absoluteAnMinusGn = absoluteAnMinusGn * (-1); //Ensures absolute value of continuation condition.
        }

    printf ("DEBUG IN: x%d, y%d, e%d, absoulteAnMinusGn=%f, a=%f, g=%f, an=%f, gn=%f\n", x,y,e,absoluteAnMinusGn,a,g,an,gn);//DEBUG CODE 

    while (absoluteAnMinusGn > e) {
        an = (a + g) / 2;
        gn = sqrt (a * g);
        a = an;
        g = gn;
        absoluteAnMinusGn = an - gn;
        if (absoluteAnMinusGn < 0) {
            absoluteAnMinusGn = absoluteAnMinusGn * (-1);
        }
    }

    printf ("DEBUG OUT: x=%d, y=%d, e=%d, absoulteAnMinusGn=%f, a=%f, g=%f, an=%f, gn=%f\n", x,y,e,absoluteAnMinusGn,a,g,an,gn);//DEBUG CODE 
    return 0;
}

然后当我用下面显示的一些样本数运行它时，我得到了结果。

C中返回大数的命令行参数

3 个答案: