多项式运算
Polynom manipulations
我想添加两个多项式并将结果设置为多项式结果我对第三个函数有疑问请提供任何指示我真的需要帮助
PS: 动态声明!看来我不太了解指针,所以如果你向我建议任何可能有助于我理解这一章的东西,那将很有帮助!
typedef struct {
int degree;
float* coeff_tab;
} Polynome;
Polynome lire_polynome()
{
Polynome p;
int i;
printf("Degree du polynome? ");
scanf("%d", &p.degree);
p.coeff_tab = (float*)malloc((unsigned long)(p.degree+1)*sizeof (float));
for(i=0; i<=p.degree; i++)
{
printf("coeff de degree %i: ", i);
scanf("%f", &p.coeff_tab[i]);
}
//free(p.coeff_tab);
return p;
}
void affiche_polynome(Polynome p)
{
int i;
if(p.degree > 1){
printf("%4.2fX^%i ", p.coeff_tab[p.degree], p.degree);
for(i=p.degree-1; i>1; i--){
printf(" + %4.2fX^%i ", p.coeff_tab[i], i);
}
printf(" + %4.2fX + %4.2f\n", p.coeff_tab[1], p.coeff_tab[0]);
} else {
printf(" + %4.2f\n", p.coeff_tab[0]);
}
}
void sommePoly(Polynome p1 , Polynome p2 , Polynome psom)
{
int degsom,deg1,deg2,i=0,j=0,k=0;
float coeffsum,coeff1,coeff2;
Polynome *P,*S ,*Psom;
P=&p1; S=&p2;
degsom=max(p1.degree,p2.degree);
psom.coeff_tab=(float*)malloc((unsigned long)(degsom+1)*sizeof (float));
Psom=&psom;
if( p1.degree == p2.degree)
{
psom.coeff_tab[k]=p1.coeff_tab[i]+p2.coeff_tab[j];
psom.degree=p1.degree;
Psom ++;
i++;j++;k++;
}
else if(p1.degree > p2.degree) {
psom.degree=p1.degree;
psom.coeff_tab[k]=p1.coeff_tab[i];
i++;k++;
}else {
psom.coeff_tab[k]=p2.coeff_tab[j];
psom.degree=p2.degree;
j++;k++;
}
}
void sommePoly(Polynome p1 , Polynome p2 , Polynome psom)
在 C 中,参数是通过复制 变量值来传递的。所以 p1
是你传递的变量的副本就是它。您需要传递指针才能修改指针后面的值。或者 return 值,就像在 Polynome lire_polynome()
.
的情况下所做的那样
通常我期望函数:
int lire_polynome(Polynome *p);
int sommePoly(Polynome *p1 , Polynome *p2 , Polynome *psom);
使用一些 return 值,例如。 int 你可以 return 负数或非零数以防出错(例如 malloc
错误)。
所以你的函数可能是这样的:
int sommePoly(Polynome *p1 , Polynome *p2 , Polynome *psom)
{
int degsom, deg1, deg2, i = 0, j = 0, k = 0;
float coeffsum, coeff1, coeff2;
degsom = max(p1->degree, p2->degree);
psom->coeff_tab = malloc((degsom + 1) * sizeof(float));
if (psom->coeff_tab == NULL) { return -1; }
if (p1->degree == p2->degree){
psom->coeff_tab[k] = p1->coeff_tab[i] + p2->coeff_tab[j];
psom->degree = p1->degree;
... // and so on replace psom. to psom-> ie. replace dot with ->
}
return 0;
}
然后你称之为传递指针到结构:
Polynome p1;
if (lire_polynome(&p1) != 0) { /* error handling */ }
Polynome p2;
if (lire_polynome(&p2) != 0) { /* error ahdling */ }
Polynome p3;
if (sommePoly(&p1, &p2, &p3) != 0) { /* error handling */ }
- 记得检查
scanf
个错误 if (scanf("%f", ...) != 1) { fprintf(stderr, "error in scanf"); exit(1); }
- 记得检查 malloc 错误。
- Don't cast result of malloc
- 在
(unsigned long)(p.degree+1)*sizeof (float)
中转换为 unsigned long 很奇怪而且很容易出错。 sizeof(...)
是 size_t 类型,这是大小表示的正确类型,将其转换为 unsigned long
是不必要的,并且在某些大数字情况下可能会导致错误。只需 malloc((p.degree + 1) * sizeof(float))
或 calloc(p->degree + 1, sizeof(float));
- 好好读一读 linux kernel coding style。
- 可以在 this thread.
中找到传递指针和修改值的更简单示例
我想这个版本的 sommePoly 就是你要找的:
void sommePoly(Polynome *p1 , Polynome *p2 , Polynome *psom)
{
int i;
Polynome *big, *sml;
if (p1->degree > p2->degree) {
big = p1;
sml = p2;
}
else {
big = p2;
sml = p1;
}
psom->degree = big->degree;
psom->coeff_tab = calloc(psom->degree + 1, sizeof(float));
for (i = 0; i < sml->degree + 1; i++)
{
psom->coeff_tab[i] = big->coeff_tab[i] + sml->coeff_tab[i];
}
for (; i < big->degree + 1; i++)
{
psom->coeff_tab[i] = big->coeff_tab[i];
}
}
提示 1:尝试使用 calloc() 而不是 malloc()。尝试分配任何 Z 大小的 N 项数组时要安全得多。
提示 2:当然,在取消引用之前总是检查 return calloc()/malloc() 的值。
说明 sommePoly 的预期用途:
void printPoly(Polynome *p)
{
int i;
for (i = p->degree; i > 0; i--) {
if (0.0 != p->coeff_tab[i])
printf("%4.2fX^%i + ", p->coeff_tab[i], i);
}
if (0.0 != p->coeff_tab[i])
printf("%4.2f\n", p->coeff_tab[0]);
}
int main(int argc, char *argv[])
{
Polynome a, b, sum;
a.degree = 5;
a.coeff_tab = calloc(a.degree + 1, sizeof(float));
a.coeff_tab[0] = 1.0;
a.coeff_tab[1] = 8.0;
a.coeff_tab[2] = -2.0;
a.coeff_tab[3] = 0.0;
a.coeff_tab[4] = 3.0;
a.coeff_tab[5] = 1.0;
b.degree = 3;
b.coeff_tab = calloc(b.degree + 1, sizeof(float));
b.coeff_tab[0] = 1.0;
b.coeff_tab[1] = -3.0;
b.coeff_tab[2] = 5.0;
b.coeff_tab[3] = 7.0;
sommePoly(&a, &b, &sum);
printPoly(&a);
printPoly(&b);
printPoly(&sum);
free(a.coeff_tab);
free(b.coeff_tab);
free(sum.coeff_tab);
return 0;
}
和输出:
$ cc -g -O0 -Wall poly.c -o poly
$ ./poly
1.00X^5 + 3.00X^4 + -2.00X^2 + 8.00X^1 + 1.00
7.00X^3 + 5.00X^2 + -3.00X^1 + 1.00
1.00X^5 + 3.00X^4 + 7.00X^3 + 3.00X^2 + 5.00X^1 + 2.00
我想添加两个多项式并将结果设置为多项式结果我对第三个函数有疑问请提供任何指示我真的需要帮助 PS: 动态声明!看来我不太了解指针,所以如果你向我建议任何可能有助于我理解这一章的东西,那将很有帮助!
typedef struct {
int degree;
float* coeff_tab;
} Polynome;
Polynome lire_polynome()
{
Polynome p;
int i;
printf("Degree du polynome? ");
scanf("%d", &p.degree);
p.coeff_tab = (float*)malloc((unsigned long)(p.degree+1)*sizeof (float));
for(i=0; i<=p.degree; i++)
{
printf("coeff de degree %i: ", i);
scanf("%f", &p.coeff_tab[i]);
}
//free(p.coeff_tab);
return p;
}
void affiche_polynome(Polynome p)
{
int i;
if(p.degree > 1){
printf("%4.2fX^%i ", p.coeff_tab[p.degree], p.degree);
for(i=p.degree-1; i>1; i--){
printf(" + %4.2fX^%i ", p.coeff_tab[i], i);
}
printf(" + %4.2fX + %4.2f\n", p.coeff_tab[1], p.coeff_tab[0]);
} else {
printf(" + %4.2f\n", p.coeff_tab[0]);
}
}
void sommePoly(Polynome p1 , Polynome p2 , Polynome psom)
{
int degsom,deg1,deg2,i=0,j=0,k=0;
float coeffsum,coeff1,coeff2;
Polynome *P,*S ,*Psom;
P=&p1; S=&p2;
degsom=max(p1.degree,p2.degree);
psom.coeff_tab=(float*)malloc((unsigned long)(degsom+1)*sizeof (float));
Psom=&psom;
if( p1.degree == p2.degree)
{
psom.coeff_tab[k]=p1.coeff_tab[i]+p2.coeff_tab[j];
psom.degree=p1.degree;
Psom ++;
i++;j++;k++;
}
else if(p1.degree > p2.degree) {
psom.degree=p1.degree;
psom.coeff_tab[k]=p1.coeff_tab[i];
i++;k++;
}else {
psom.coeff_tab[k]=p2.coeff_tab[j];
psom.degree=p2.degree;
j++;k++;
}
}
void sommePoly(Polynome p1 , Polynome p2 , Polynome psom)
在 C 中,参数是通过复制 变量值来传递的。所以 p1
是你传递的变量的副本就是它。您需要传递指针才能修改指针后面的值。或者 return 值,就像在 Polynome lire_polynome()
.
通常我期望函数:
int lire_polynome(Polynome *p);
int sommePoly(Polynome *p1 , Polynome *p2 , Polynome *psom);
使用一些 return 值,例如。 int 你可以 return 负数或非零数以防出错(例如 malloc
错误)。
所以你的函数可能是这样的:
int sommePoly(Polynome *p1 , Polynome *p2 , Polynome *psom)
{
int degsom, deg1, deg2, i = 0, j = 0, k = 0;
float coeffsum, coeff1, coeff2;
degsom = max(p1->degree, p2->degree);
psom->coeff_tab = malloc((degsom + 1) * sizeof(float));
if (psom->coeff_tab == NULL) { return -1; }
if (p1->degree == p2->degree){
psom->coeff_tab[k] = p1->coeff_tab[i] + p2->coeff_tab[j];
psom->degree = p1->degree;
... // and so on replace psom. to psom-> ie. replace dot with ->
}
return 0;
}
然后你称之为传递指针到结构:
Polynome p1;
if (lire_polynome(&p1) != 0) { /* error handling */ }
Polynome p2;
if (lire_polynome(&p2) != 0) { /* error ahdling */ }
Polynome p3;
if (sommePoly(&p1, &p2, &p3) != 0) { /* error handling */ }
- 记得检查
scanf
个错误if (scanf("%f", ...) != 1) { fprintf(stderr, "error in scanf"); exit(1); }
- 记得检查 malloc 错误。
- Don't cast result of malloc
- 在
(unsigned long)(p.degree+1)*sizeof (float)
中转换为 unsigned long 很奇怪而且很容易出错。sizeof(...)
是 size_t 类型,这是大小表示的正确类型,将其转换为unsigned long
是不必要的,并且在某些大数字情况下可能会导致错误。只需malloc((p.degree + 1) * sizeof(float))
或calloc(p->degree + 1, sizeof(float));
- 好好读一读 linux kernel coding style。
- 可以在 this thread. 中找到传递指针和修改值的更简单示例
我想这个版本的 sommePoly 就是你要找的:
void sommePoly(Polynome *p1 , Polynome *p2 , Polynome *psom)
{
int i;
Polynome *big, *sml;
if (p1->degree > p2->degree) {
big = p1;
sml = p2;
}
else {
big = p2;
sml = p1;
}
psom->degree = big->degree;
psom->coeff_tab = calloc(psom->degree + 1, sizeof(float));
for (i = 0; i < sml->degree + 1; i++)
{
psom->coeff_tab[i] = big->coeff_tab[i] + sml->coeff_tab[i];
}
for (; i < big->degree + 1; i++)
{
psom->coeff_tab[i] = big->coeff_tab[i];
}
}
提示 1:尝试使用 calloc() 而不是 malloc()。尝试分配任何 Z 大小的 N 项数组时要安全得多。 提示 2:当然,在取消引用之前总是检查 return calloc()/malloc() 的值。
说明 sommePoly 的预期用途:
void printPoly(Polynome *p)
{
int i;
for (i = p->degree; i > 0; i--) {
if (0.0 != p->coeff_tab[i])
printf("%4.2fX^%i + ", p->coeff_tab[i], i);
}
if (0.0 != p->coeff_tab[i])
printf("%4.2f\n", p->coeff_tab[0]);
}
int main(int argc, char *argv[])
{
Polynome a, b, sum;
a.degree = 5;
a.coeff_tab = calloc(a.degree + 1, sizeof(float));
a.coeff_tab[0] = 1.0;
a.coeff_tab[1] = 8.0;
a.coeff_tab[2] = -2.0;
a.coeff_tab[3] = 0.0;
a.coeff_tab[4] = 3.0;
a.coeff_tab[5] = 1.0;
b.degree = 3;
b.coeff_tab = calloc(b.degree + 1, sizeof(float));
b.coeff_tab[0] = 1.0;
b.coeff_tab[1] = -3.0;
b.coeff_tab[2] = 5.0;
b.coeff_tab[3] = 7.0;
sommePoly(&a, &b, &sum);
printPoly(&a);
printPoly(&b);
printPoly(&sum);
free(a.coeff_tab);
free(b.coeff_tab);
free(sum.coeff_tab);
return 0;
}
和输出:
$ cc -g -O0 -Wall poly.c -o poly
$ ./poly
1.00X^5 + 3.00X^4 + -2.00X^2 + 8.00X^1 + 1.00
7.00X^3 + 5.00X^2 + -3.00X^1 + 1.00
1.00X^5 + 3.00X^4 + 7.00X^3 + 3.00X^2 + 5.00X^1 + 2.00