C program that can expand quadratics

问题

I want to have a C program that allows me to input (x+1)(x+3) and other stuff like that, including x^2. So far I have a very complex system using linked lists but I think that there should be an easier solution. The output from input, (x+1)(x+3) would be x^2+4x+3 printed out.

So far I'm passing around a struct _term with an int, char and int for coefficient,pro numeral and power. so 2x^4 would be saved as |2|'x'|3|.

I should also mention that I'm only 16, and still in high school.

回答1:

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

#define BUFFSIZE 128

typedef struct _exp { //Integral expression
    int n; //Max of order
    int *k;//coefficient list(array)# k[0] as constant, k[n] as x^n
} Exp;


void print(Exp *exp);
const char *extract(const char *str, char *outbuff);
Exp *makeExp(const char *str);
void freeExp(Exp *exp);
Exp *convert(const char *str);
Exp *mul(Exp *a, Exp *b);
Exp *calc(const char *expStr);

int main(){
    char buff[BUFFSIZE];
    Exp *exp;
    printf("input expression Eg.(x+1)(x+3)\n>");
    fgets(buff, sizeof(buff)/sizeof(char), stdin);
    exp=calc(buff);
    print(exp);
    freeExp(exp);

    return 0;
}

void print(Exp *exp){
    int i, n = exp->n;
    for(i=n;i>=0;--i){
        int k = exp->k[i];
        if(k==0)continue;
        if(k<0)
            printf("-");
        if(k>0 && i!=n)
            printf("+");
        if(k<0) k*=-1;//k=|k|
        if(i!=0){
            if(k!=1)
                printf("%dx", k);
            else
                printf("x");
        }
        if(i>1)
            printf("^%d", i);
        if(i==0)
            printf("%d", k);
    }
    printf("\n");
}


const char *extract(const char *str, char *outbuff){
//pull out in parenthesis to outbuff, and remove space
    while(isspace(*str))
        ++str;
    if(*str=='\0') return NULL;
    while(*str!='(')++str;
    while(*++str!=')')
        if(!isspace(*str))
            *outbuff++=*str;
    *outbuff='\0';
    return ++str;
}

Exp *makeExp(const char *str){
    Exp *exp;
    char buff[BUFFSIZE];
    exp = (Exp*)malloc(sizeof(Exp));
    exp->n = 0;
    exp->k = (int*)calloc(1,sizeof(int));
    while(*str){
        int k,n;
        char *p;
        k=(int)strtol(str, &p, 10);
        if(k==0){//never constant zero e.g(x+0)is invalid, valid E.g (x)
            if(p[0]=='-'){
                k = -1;
                ++p;
            } else if(p[0]=='+') {
                k = 1;
                ++p;
            } else 
                k = 1;
        }
        if(*p=='\0'){
            exp->k[0]=k;//constant part
            break;
        }
        if(*p=='x'){
            n = (p[1]=='^') ? (int)strtol(&p[2], &p, 10):1;
            if(exp->n < n){
                int i;
                exp->k = (int*)realloc(exp->k, (n+1)*sizeof(int));
                for(i=n;i>exp->n;--i)
                    exp->k[i]=0;
                exp->n = n;
            }
            exp->k[n] = k;
            if(*p=='\0' || p[1]=='\0')break;
            else{
                str = (n!=1)? p:p+1;
            }
        } else {
            exp->k[0]=k;
            str = p;
        }
    }
    return exp;
}

void freeExp(Exp *exp){
    free(exp->k);
    free(exp);
}

Exp *convert(const char *str){
    Exp *exp;
    exp=makeExp(str);
    return exp;
}

Exp *mul(Exp *a, Exp *b){
    Exp *ret;
    int i,j;
    ret=(Exp*)malloc(sizeof(Exp));
    ret->n = a->n + b->n;
    ret->k = (int*)calloc(ret->n + 1, sizeof(int));
    for(i=0;i<=a->n;++i){
        for(j=0;j<=b->n;++j){
            ret->k[i+j] += a->k[i] * b->k[j];
        }
    }
    return ret;
}

Exp *calc(const char *expStr){
    Exp *exp;
    char buff[BUFFSIZE];
    const char *p=expStr;
    exp = (Exp*)malloc(sizeof(Exp));
    exp->n = 0;
    exp->k = (int*)malloc(sizeof(int));
   *exp->k = 1;
    while(NULL!=(p=extract(p, buff))){
        Exp *e,*wk;
        wk=exp;
        e=convert(buff);
        exp=mul(wk, e);
        freeExp(wk);
        freeExp(e);
    }
    return exp;
}

回答2:

You'll need to write a parser for the expression, then walk the syntax tree of it, then do some logic to make the multlpications, maybe to merge the additions, then output it. You may find this link useful on how to write an expression parser in C.

来源：https://stackoverflow.com/questions/11684058/c-program-that-can-expand-quadratics