Question

我正在建立一个定义多项式的结构，即它包含两个变量：
-int degree包含多项式的阶数
-int * coeff = (int*) malloc (degree * sizeof(int))包含所有系数

我还定义了一个函数new_polynome()，该函数接受一个度数，并返回一个指向该结构的指针，该结构保存该度数的多项式并将其所有系数都设置为1；

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

// 1

typedef struct 
{
    int degree;
    int * coeff = (int *) malloc (degree * sizeof(int));
} polynome;


// 2

polynome * new_polynome(int n)
{
    polynome * newest_polynome = (polynome *) malloc (sizeof(polynome));
    for(int i = 0; i < n; i++)
        newest_polynome->coeff[i] = 1;

    return newest_polynome;
}

int main(void)
{
    polynome * new_polynome = (polynome *) malloc (sizeof(polynome));
    new_polynome = new_polynome(5);

    for(int i = 0; i < 5; i++)
        printf("%d", new_polynome->coeff[i]);

    free(new_polynome->coeff);
    return 0;
}

但是，当我尝试打印其系数时，出现以下错误。有正确的方法吗？我不明白该错误信息。 如何打印其系数？ 我的错误如下：
TD_polynomes.c:9:17: error: expected ‘:’, ‘,’, ‘;’, ‘}’ or ‘__attribute__’ before ‘=’ token int * coeff = (int *) malloc (degree * sizeof(int)); ^ TD_polynomes.c: In function ‘new_polynome’: TD_polynomes.c:19:24: error: ‘polynome {aka struct <anonymous>}’ has no member named ‘coeff’ newest_polynome->coeff[i] = 1; ^~ TD_polynomes.c: In function ‘main’: TD_polynomes.c:27:20: error: called object ‘new_polynome’ is not a function or function pointer new_polynome = new_polynome(5); ^~~~~~~~~~~~ TD_polynomes.c:26:16: note: declared here polynome * new_polynome = (polynome *) malloc (sizeof(polynome)); ^~~~~~~~~~~~ TD_polynomes.c:30:34: error: ‘polynome {aka struct <anonymous>}’ has no member named ‘coeff’ printf("%d", new_polynome->coeff[i]); ^~ TD_polynomes.c:32:22: error: ‘polynome {aka struct <anonymous>}’ has no member named ‘coeff’ free(new_polynome->coeff); ^~

Answer 1

后面的错误通常是更早的结果，这是常见的情况。您对多义词的定义有误。结果是它没有名为coeff的成员。

您不能以您的方式初始化结构。删除结构中的malloc调用，您将看到其他一些错误将神奇地消失。

这行：new_polynome = new_polynome(5);没有意义。似乎您正在尝试为5个coeff分配空间，但是那完全是错误的。您应该喜欢这个`new_polynome-> coeff = malloc（5 * sizeof（*（new_polynome-> coeff））

还有don't cast malloc

您应该移动一些代码，例如分配coeffs。那应该放在new_polynome函数中，我将其重命名为create_polynome。

工作（和固定）代码：

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

typedef struct
{
    int degree;
    int *coeff; // No init here. You're just defining a type here.
} polynome;


polynome *create_polynome(int n)
{
    // A polynomial of degree 0 is a constant
    const int N = n+1;

   // You need to set the degree
    new_polynome->degree = n;

    polynome *new_polynome = malloc(sizeof(*new_polynome));
    new_polynome->coeff = malloc(N * sizeof(*(new_polynome->coeff)));
    for(int i = 0; i < N; i++)
        new_polynome->coeff[i] = 1;

    return new_polynome;
}

int main(void)
{
    polynome *new_polynome = create_polynome(5);
    // Looping to degree makes much more sense
    for(int i = 0; i < new_polynome->degree+1; i++)
        printf("%d", new_polynome->coeff[i]);

    free(new_polynome->coeff);
    // Not necessary to free in the end, but if you want to do it,
    // do it properly and free everything. One malloc for each free and
    // vice versa
    free(new_polynome);
    // No need  to return 0 since that's the default in main
}

我还做了一些很好的（主观地说）修改，例如在解引用器*和标识符之间没有空格。还消除了功能和感觉之间的空间。另外，我使用对象的大小而不是类型。

作为一个额外的奖励，该函数派生自：

polynome *derivate_polynom(polynome *pol)
{
    polynome * derivate = create_polynome(pol->degree - 1);
    for(int i = 0; i < derivate->degree + 1; i++)
        derivate->coeff[i] = (i+1) * pol->coeff[i+1];
    return derivate;
}

Answer 2

1）int * coeff = (int *) malloc (degree * sizeof(int));在结构定义中的语法无效：

typedef struct 
{
    int degree;
    int * coeff = (int *) malloc (degree * sizeof(int));
} polynome;

应该是：

typedef struct 
{
    int degree;
    int * coeff;
} polynome;

2）这是错误的：

polynome * new_polynome = (polynome *) malloc (sizeof(polynome));
new_polynome = new_polynome(5);

您正在将malloc的结果分配给new_polynome，然后立即用返回值new_polynome(5).覆盖它，这是内存泄漏。

3）我想您可能想为N+1整数而不是N分配空间。零级多项式是一个常数。

Answer 3

这是无效的语法；你可以写

typedef struct 
{
    int degree;
    int coeff[];
} polynome;

polynome *new_polynome(int n)
{
    polynome *p;

    p = malloc(sizeof *p + n * sizeof p->coeff[0]);
    for(int i = 0; i < n; i++)
        p->coeff[i] = 1;

    return p;
}

而不是。也许，当n是不受信任的用户输入时，检查是否溢出。

在结构内分配动态数组

3 个答案: