c - 缩短的 for/foreach 循环

Question

我在网上找到了一段代码，其中有一行我的编译器无法运行。我也不明白怎么才能正常写。

#define _sum_k(a, b, c, s) { s = 0; foreach(k, a, b) s+= c; }

foreach(k, a, b)意味着 for (int k = a; k < b; k++)

我该如何像普通的 for 循环一样编写它？

代码用于 LU 分解，代码如下:

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

#define foreach(a, b, c) for (int a = b; a < c; a++)
#define for_i foreach(i, 0, n)
#define for_j foreach(j, 0, n)
#define for_k foreach(k, 0, n)
#define for_ij for_i for_j
#define for_ijk for_ij for_k
#define _dim int n
#define _swap(x, y) { typeof(x) tmp = x; x = y; y = tmp; }
#define _sum_k(a, b, c, s) { s = 0; foreach(k, a, b) s+= c; }

typedef double **mat;

#define _zero(a) mat_zero(a, n)
void mat_zero(mat x, int n) {
    for_ij
            x[i][j] = 0;
}

#define _new(a) a = mat_new(n)
mat mat_new(_dim) {
    mat x = malloc(sizeof(double*) * n);
    x[0] = malloc(sizeof(double) * n * n);

    for_i
        x[i] = x[0] + n * i;
    _zero(x);

    return x;
}

#define _copy(a) mat_copy(a, n)
mat mat_copy(void *s, _dim) {
    mat x = mat_new(n);
    for_ij
            x[i][j] = ((double(*)[n]) s)[i][j];
    return x;
}

#define _del(x) mat_del(x)
void mat_del(mat x) {
    free(x[0]);
    free(x);
}

#define _QUOT(x) #x
#define QUOTE(x) _QUOT(x)
#define _show(a) printf(QUOTE(a)" =");mat_show(a, 0, n)
void mat_show(mat x, char *fmt, _dim) {
    if (!fmt)
        fmt = "%8.4g";
    for_i {
        printf(i ? "      " : " [ ");
        for_j {
            printf(fmt, x[i][j]);
            printf(j < n - 1 ? "  " : i == n - 1 ? " ]\n" : "\n");
        }
    }
}

#define _mul(a, b) mat_mul(a, b, n)
mat mat_mul(mat a, mat b, _dim) {
    mat c = _new(c);
    for_ijk
                c[i][j] += a[i][k] * b[k][j];
    return c;
}

#define _pivot(a, b) mat_pivot(a, b, n)
void mat_pivot(mat a, mat p, _dim) {
    for_ij
        {
            p[i][j] = (i == j);
        }
    for_i {
        int max_j = i;
        foreach(j, i, n)
            if (fabs(a[j][i]) > fabs(a[max_j][i]))
                max_j = j;

        if (max_j != i)
            for_k {
                _swap(p[i][k], p[max_j][k]);
            }
    }
}

#define _LU(a, l, u, p) mat_LU(a, l, u, p, n)
void mat_LU(mat A, mat L, mat U, mat P, _dim) {
    _zero(L);
    _zero(U);
    _pivot(A, P);

    mat Aprime = _mul(P, A);

    for_i {
        L[i][i] = 1;
    }
    for_ij
        {
            double s;
            if (j <= i) {
                _sum_k(0, j, L[j][k] * U[k][i], s)
                U[j][i] = Aprime[j][i] - s;
            }
            if (j >= i) {
                _sum_k(0, i, L[j][k] * U[k][i], s);
                L[j][i] = (Aprime[j][i] - s) / U[i][i];
            }
        }

    _del(Aprime);
}

double A3[][3] = { { 1, 3, 5 }, { 2, 4, 7 }, { 1, 1, 0 } };
double A4[][4] = { { 11, 9, 24, 2 }, { 1, 5, 2, 6 }, { 3, 17, 18, 1 }, { 2, 5,
        7, 1 } };

int main() {
    int n = 3;
    mat A, L, P, U;

    _new(L);
    _new(P);
    _new(U);
    A = _copy(A3);
    _LU(A, L, U, P);
    _show(A);
    _show(L);
    _show(U);
    _show(P);
    _del(A);
    _del(L);
    _del(U);
    _del(P);

    printf("\n");

    n = 4;

    _new(L);
    _new(P);
    _new(U);
    A = _copy(A4);
    _LU(A, L, U, P);
    _show(A);
    _show(L);
    _show(U);
    _show(P);
    _del(A);
    _del(L);
    _del(U);
    _del(P);

    return 0;
}

编辑:我替换了每个宏，包括有问题的宏(使用 s = c * (b-a) * ( b > a); )，但它崩溃了，所以也许这不是它应该的样子。

Answer 1

预处理器只是执行“搜索和替换”。

#define foreach(a, b, c) for (int a = b; a < c; a++)字面意思是foreach(x, y, z)将替换为 for (int x = y; x < z; x++) 。等等。 它对每个 #define 都这样做是字面意思，直到没有更多的替代品为止。

您可以简单地通过预处理器运行您的程序并查看剩下的内容:

cpp myfile.c > myfile_preprocessed.c

// #include's omitted

typedef double **mat;

void mat_zero(mat x, int n) {
    for (int i = 0; i < n; i++) for (int j = 0; j < n; j++)
            x[i][j] = 0;
}

mat mat_new(int n) {
    mat x = malloc(sizeof(double*) * n);
    x[0] = malloc(sizeof(double) * n * n);

    for (int i = 0; i < n; i++)
        x[i] = x[0] + n * i;
    mat_zero(x, n);

    return x;
}

mat mat_copy(void *s, int n) {
    mat x = mat_new(n);
    for (int i = 0; i < n; i++) for (int j = 0; j < n; j++)
            x[i][j] = ((double(*)[n]) s)[i][j];
    return x;
}

void mat_del(mat x) {
    free(x[0]);
    free(x);
}

void mat_show(mat x, char *fmt, int n) {
    if (!fmt)
        fmt = "%8.4g";
    for (int i = 0; i < n; i++) {
        printf(i ? "      " : " [ ");
        for (int j = 0; j < n; j++) {
            printf(fmt, x[i][j]);
            printf(j < n - 1 ? "  " : i == n - 1 ? " ]\n" : "\n");
        }
    }
}


mat mat_mul(mat a, mat b, int n) {
    mat c = c = mat_new(n);
    for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) for (int k = 0; k < n; k++)
                c[i][j] += a[i][k] * b[k][j];
    return c;
}


void mat_pivot(mat a, mat p, int n) {
    for (int i = 0; i < n; i++) for (int j = 0; j < n; j++)
        {
            p[i][j] = (i == j);
        }
    for (int i = 0; i < n; i++) {
        int max_j = i;
        for (int j = i; j < n; j++)
            if (fabs(a[j][i]) > fabs(a[max_j][i]))
                max_j = j;

        if (max_j != i)
            for (int k = 0; k < n; k++) {
                { typeof(p[i][k]) tmp = p[i][k]; p[i][k] = p[max_j][k]; p[max_j][k] = tmp; };
            }
    }
}


void mat_LU(mat A, mat L, mat U, mat P, int n) {
    mat_zero(L, n);
    mat_zero(U, n);
    mat_pivot(A, P, n);

    mat Aprime = mat_mul(P, A, n);

    for (int i = 0; i < n; i++) {
        L[i][i] = 1;
    }
    for (int i = 0; i < n; i++) for (int j = 0; j < n; j++)
        {
            double s;
            if (j <= i) {
                { s = 0; for (int k = 0; k < j; k++) s+= L[j][k] * U[k][i]; }
                U[j][i] = Aprime[j][i] - s;
            }
            if (j >= i) {
                { s = 0; for (int k = 0; k < i; k++) s+= L[j][k] * U[k][i]; };
                L[j][i] = (Aprime[j][i] - s) / U[i][i];
            }
        }

    mat_del(Aprime);
}

double A3[][3] = { { 1, 3, 5 }, { 2, 4, 7 }, { 1, 1, 0 } };
double A4[][4] = { { 11, 9, 24, 2 }, { 1, 5, 2, 6 }, { 3, 17, 18, 1 }, { 2, 5,
        7, 1 } };

int main() {
    int n = 3;
    mat A, L, P, U;

    L = mat_new(n);
    P = mat_new(n);
    U = mat_new(n);
    A = mat_copy(A3, n);
    mat_LU(A, L, U, P, n);
    printf("A"" =");mat_show(A, 0, n);
    printf("L"" =");mat_show(L, 0, n);
    printf("U"" =");mat_show(U, 0, n);
    printf("P"" =");mat_show(P, 0, n);
    mat_del(A);
    mat_del(L);
    mat_del(U);
    mat_del(P);

    printf("\n");

    n = 4;

    L = mat_new(n);
    P = mat_new(n);
    U = mat_new(n);
    A = mat_copy(A4, n);
    mat_LU(A, L, U, P, n);
    printf("A"" =");mat_show(A, 0, n);
    printf("L"" =");mat_show(L, 0, n);
    printf("U"" =");mat_show(U, 0, n);
    printf("P"" =");mat_show(P, 0, n);
    mat_del(A);
    mat_del(L);
    mat_del(U);
    mat_del(P);

    return 0;
}

请注意，不要用函数调用替换 for 循环的主体，生成的代码可能会变慢，这取决于优化器的质量和设置。