c - 两个矩阵大小超过800 * 800时出现段错误

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Segmentation fault on large array sizes
（5个答案）
我试图为整数数据类型编写一个向量乘加“axpy”算法的超级简单C程序。程序输出执行时间以测量机器的性能。矩阵由随机数填充。

int benchmark(void) {
    int N;      /* The matrix size, controlled by user input */
    int r, c;   /* Row and Column number */
    int random; /* Random number to fill the matix */
    int a = rand() % 20; /* Scale number to multiply x matrix */

    printf("Enter the size(N*N) of the matrices(Maximum 1,000,000)\n");
    scanf("%d", &N);

    if (N > 1000000) {
        fprintf(stderr, "Size of matrix is too large!\n");
        return 0;
    }

    /* Initialize and fill the matrix x and y */
    int xMatrix[N][N], yMatrix[N][N], resultMatrix[N][N];

    /* Compute time */
    clock_t t;

    t = clock();

    for (r = 0; r < N; r++) {
        for (c = 0; c < N; c++) {
            random = rand() % 100;
            xMatrix[r][c] = a * random; /* Multiply matrix x with random value a */
        }
    }
    for (r = 0; r < N; r++) {
        for (c = 0; c < N; c++) {
            int random = rand() % 100;
            yMatrix[r][c] = random;
        }
    }

    /* Add two matrix together */
    for (r = 0; r < N; r++) {
        for (c = 0; c < N; c++) {
            resultMatrix[r][c] = xMatrix[r][c] + yMatrix[r][c];
        }
    }

    t = clock() - t;

    double timeTaken = ((double)t) / CLOCKS_PER_SEC;
    printf("\n -> Total time : %f seconds\n", timeTaken);
    printf("\n -> Vector length : %d", N * N);

}

用户控制矩阵的大小。
当N的值小于800时，程序运行良好。

最佳答案

使用自动存储（在堆栈上）分配的对象太大，会出现未定义的行为，更确切地说是堆栈溢出。
相反，您应该从堆中分配对象：

    /* Initialize and fill the matix x and y */
    int (*xMatrix)[N] = malloc(N * sizeof(*xMatrix));
    int (*yMatrix)[N] = malloc(N * sizeof(*yMatrix));
    int (*resultMatrix)[N] = malloc(N * sizeof(*resultMatrix));

并验证malloc()返回的指针都不是NULL。
下面是修改后的代码：

int benchmark(void) {
    int N;       /* The matrix size, controlled by user input */
    int r, c;    /* Row and Column number */
    int random;  /* Random number to fill the matix */
    int a = rand() % 20; /* Scale number to multiply x matrix */

    printf("Enter the size(N*N) of the matrices (Maximum 1,000,000)\n");
    if (scanf("%d", &N) != 1) {
        fprintf(stderr, "Input error!\n");
        return 0;
    }

    if (N > 1000000) {
        fprintf(stderr, "Matrix size is too large!\n");
        return 0;
    }

    /* Initialize and fill the matrix x and y */
    int (*xMatrix)[N] = malloc(N * sizeof(*xMatrix));
    int (*yMatrix)[N] = malloc(N * sizeof(*yMatrix));
    int (*resultMatrix)[N] = malloc(N * sizeof(*resultMatrix));

    if (xMatrix == NULL || yMatrix == NULL || resultMatrix == NULL) {
        fprintf(stderr, "Memory allocation failed!\n");
        free(xMatrix);
        free(yMatrix);
        free(resultMatrix);
        return 0;
    }

    /* Compute time */
    clock_t t = clock();

    for (r = 0; r < N; r++) {
        for (c = 0; c < N; c++) {
            random = rand() % 100;
            xMatrix[r][c] = a * random; /* Multiply matrix x with random value a */
        }
    }
    for (r = 0; r < N; r++) {
        for (c = 0; c < N; c++) {
            random = rand() % 100;
            yMatrix[r][c] = random;
        }
    }

    /* Add two matrix together */
    for (r = 0; r < N; r++) {
        for (c = 0; c < N; c++) {
            resultMatrix[r][c] = xMatrix[r][c] + yMatrix[r][c];
        }
    }

    t = clock() - t;

    double timeTaken = ((double)t) / CLOCKS_PER_SEC;
    printf("\n -> Total time : %f seconds\n", timeTaken);
    printf("\n -> Vector length : %lld", (long long)N * N);

    free(xMatrix);
    free(yMatrix);
    free(resultMatrix);
    return 0;
}

但是请注意，您的计算非常简单，大部分时间可能都花在rand()函数中。

关于c - 两个矩阵大小超过800 * 800时出现段错误，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/46616044/