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这是我得到的代码。它编译得很好，我根本看不出可能出了什么问题：

    // This program takes a quadratic from the user, and prints the solution(s) if they exist. 

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


//Make sign function for switch test - because in the switch expression itself, C refused to believe x/|x| is an integer...
int sign(float f){
    printf("\n\nSIGN CALL OK\n\n");
    int sign=f/cabs(f);
    return sign;
}

// Define quadratic structure 
typedef struct quadratic{
    float a, b, c;
    float discriminant;
    float real_root_1;
    float real_root_2;
    float complex_root;
} Quadratic;

// 'Redefine' malloc to also check allocation of memory when called.

Quadratic* xmalloc(size_t n){
    printf("\n\nXMALLOC CALL OK\n\n");
    Quadratic* p = malloc(n);
    if (p == NULL) {
        printf("\n ERROR: Unable to allocate memory! \n");
        exit(1);
    }
    return p;
}

// newquadratic lets the user define the quadratic to be solved, and returns the pointer to its structure. 

Quadratic* newquadratic() {
    Quadratic* q = xmalloc(sizeof *q);
    printf("\n Please enter the coefficients of the quadratic separated by spaces: ");
    scanf("%g, %g, %g", q->a, q->b, q->c);
    printf("\n\nDATA ACCEPTED\n\n");
    return q;
}

// solve takes the existing data from the quadratics structure and defines the remaining quantities, depending on which
// case we get for the 'sign of the discriminant'.

int solve(Quadratic eqn){
    printf("\n\nSOLVE CALL OK\n\n");
    eqn.discriminant = (eqn.b*eqn.b - 4*eqn.a*eqn.c);

    switch (sign(eqn.discriminant)) {
        case -1:
            eqn.real_root_1 = (-eqn.b+sqrt(eqn.discriminant))/(2*eqn.a);
            eqn.real_root_2 = (-eqn.b-sqrt(eqn.discriminant))/(2*eqn.a);
            break;
        case 0:
            eqn.real_root_1 = (-eqn.b+sqrt(eqn.discriminant))/(2*eqn.a);
            break;
        case 1:
            eqn.complex_root = (-eqn.b+sqrt(eqn.discriminant))/(2*eqn.a);
            break;
    }
    return sign(eqn.discriminant);
}

//main also uses sign of the discriminant (as returned by solve) to decide how to behave appropriately for the case it receives.

int main () {

    Quadratic* eqn=newquadratic();

    printf("\n\n GET QUADRATIC OK\n\n");

    switch (solve(*eqn)) {
        case -1:
            printf("\n\n We have two real solutions, %g and %g", eqn->real_root_1, eqn->real_root_2);
            break;
        case 0:
            printf("\n\n We have one repeated real solution, %g", eqn->real_root_1);
            break;
        case 1:
            printf("\n\n We have two complex solutions, %g%+gi and %g%-gi", creal(eqn->complex_root),cimag(eqn->complex_root),creal(eqn->complex_root),(-1)*cimag(eqn->complex_root));
            break;
    }

        return 0;
}

xmalloc 正确调用，我们进入用户输入 3 个系数的阶段。一旦返回返回，我们就会得到错误。

4

1 回答 1

2

问题出在你的函数 newquadratic 中。

scanf 函数期望得到它应该将结果放入其中的变量的地址，而实际上您已经将变量保存的值传递给它。

要修复，请像这样传递 a、b 和 c 的地址：（添加&字符）

Quadratic* newquadratic() 
{
    Quadratic* q = xmalloc(sizeof *q);
    printf("\n Please enter the coefficients of the quadratic separated by spaces: ");
    scanf("%g, %g, %g", &q->a, &q->b, &q->c);
    printf("\n\nDATA ACCEPTED\n\n");
    return q;
}
于 2013-11-13T16:54:48.240 回答