Question

int     *rrange(int start, int end);

我被问到，编写一个函数并用malloc分配一个整数数组，用连续填充它从结束开始到结束的值（包括开始和结束！），然后返回指向数组第一个值的指针。这些函数适用于小数字，当用大数字测试时，我得到seg fault。而且我认为seg fault因为达到int的限制而得到/*- With (1, 3) you will return an array containing 3, 2 and 1 - With (-1, 2) you will return an array containing 2, 1, 0 and -1. - With (0, 0) you will return an array containing 0. - With (0, -3) you will return an array containing -3, -2, -1 and 0.*/ #include <stdlib.h> int *rrange(int start, int end) { int *range; int i; i = 0; range = (int *)malloc(sizeof(int *)); if (end <= start) { while (end <= start) range[i++] = end++; } else { while (end >= start) range[i++] = end--; } return (range); } = Test 1 =================================================== $> ./pw53y11cbachacu14eue5cab $> diff -U 3 user_output_test1 test1.output | cat -e Diff OK :D = Test 2 =================================================== $> ./o1jrm4t3vqengizvj1tlwab4 "21" "2313" "12" $> diff -U 3 user_output_test2 test2.output | cat -e Diff OK :D = Test 3 =================================================== $> ./usl3i1tc1xv9tr1gs9n5x5vr "2147483647" "2147483640" "7" $> diff -U 3 user_output_test3 test3.output | cat -e --- user_output_test3 2016-06-08 16:26:16.000000000 +0200$ +++ test3.output 2016-06-08 16:26:16.000000000 +0200$ @@ -1,8 +1,8 @@$ -0$ -0$ -0$ -0$ -0$ -0$ -0$ +2147483640$ +2147483641$ +2147483642$ +2147483643$ +2147483644$ +2147483645$ +2147483646$ $ Diff KO :( Grade: 0 = Final grade: 0 ===============================================================。我该如何解决这个问题？。

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Answer 1

您的问题来自malloc：

而不是

range = (int *)malloc(sizeof(int *));

你应该写：

range = malloc(nb_of_integer_in_range * sizeof (int));

由您来计算nb_of_integer_in_range，例如：

int nb_of_integer_in_range;
if (end > start)
{
    nb_of_integer_in_range = end - start + 1;
}
else ...

Answer 2

由于错误的内存分配而发生了段错误

range = (int *)malloc(sizeof(int *));

这会分配一个整数指针大小的内存。但要为数字分配内存，您需要以这种方式为每个数字创建所需的内存量：

range = malloc(no_of_elements*sizeof(int));

注意：使用n作为全局变量，以便您可以知道其大小甚至在函数（）中打印数组

我修改了你的功能

int *rrange(int start, int end)
{
    int *range;
    int i;

    if(start>end)
        return rrange(end,start); //so that start is always <end

    n=end-start+1; // n globally declared 

    range=malloc(n*sizeof(int)); //casting is not required
    if(range==NULL)
    {// check if memory was successfully allocated if not exit
        printf("fail");
            exit(1);
    }
    for(i=0;i<n;i++,start++)
    {
        range[i]=start;
    }

    return range;
}

你现在可以把你的主要写成：

int main()
{
    int x,y,*r,i;
    scanf("%d%d",&x,&y);
    r=rrange(x,y);
    for(i=0;i<n;i++)
    {
        printf("%d,",r[i]);
    }
    return 0;
}

Answer 3

我很惊讶这适用于任何大于2的范围。你正在分配8B的存储空间（假设你运行的是64b机器），或者通常是指针大小。你想要的是一个malloc表达式，它取决于start和end之间的差异（给你数组大小）和 sizeof(int)（给你数组）元素大小）。

不相关的注意事项：确保在停止使用阵列时释放存储空间，否则您的程序将像筛子一样泄漏。

Answer 4

malloc的论点应为number_of_integers * sizeof(int)。例如，number_of_integers的值可以计算为abs(start-end)+1，但我会在malloc - if内执行public ValueFrequency(ValueFrequency<T> valueFrequency) { if (valueFrequency.Value is ICloneable) Value = (T)((ICloneable)valueFrequency.Value).Clone(); else Value = valueFrequency.Value; _frequency = valueFrequency.Frequency; } - 开头和结尾之间的关系众所周知。

Answer 5

以下代码：

干净地编译
执行所需的功能
正确处理错误情况

start

end

可能无效，但这不属于条件

现在是代码：

/*- With (1, 3) you will return an array containing 3, 2 and 1
- With (-1, 2) you will return an array containing 2, 1, 0 and -1.
- With (0, 0) you will return an array containing 0.
- With (0, -3) you will return an array containing -3, -2, -1 and 0.*/

#include <stdlib.h> // exit(), EXIT_FAILURE
#include <stdio.h>  // perror()
#include <math.h>   // abs()

int *rrange(int start, int end)
{
    int *range = NULL;

    if( NULL == (range = malloc( sizeof(int) *((size_t)abs(end - start) +1) ) ) )
    { // then malloc failed
        perror( "malloc failed" );
        exit( EXIT_FAILURE );
    }

    for( int index = 0; index < abs(end-start)+1; index++ )
    {
        range[ index ] = start+index;
    }

    return (range);
} // end function: rrange

int指针函数seg在c中有大数字的错误

5 个答案: