返回一个int数组的Printf函数 [英] Printf function who returning an array of int

问题描述

我正在尝试此练习，但是我不知道如何在main中打印函数.练习:1)编写一个函数，该函数返回一个int选项卡，其中的所有值都在min和max之间

I'm trying this exercice but I don't know how to printf my function in main. Exercice: 1) Write a function who returning an int tab with all values between min and max

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

int *ft_range(int min, int max)
{
    int len;
    int *tab;

    len = min;
    while (len < max)
        len++;

    tab = (int *)malloc(sizeof(*tab) * len + 1);

    while (min < max)
    {
        *tab = min;
        min++;
    }
    return(tab);
}

int main()
{
    ft_range(0, 10);
    return(0);
}

推荐答案

返回一个int选项卡，其中所有值都在最小值和最大值之间

returning an int tab with all values between min and max

根据之间"的概念，是否应包含最终值是一个悬而未决的问题.鉴于OP在 sizeof(* tab)* len +1 中的误编码+1，我认为应该包括两端.

Depending on the idea of "between", it is an open question if the end values should be included. Given OP's mis-coded +1 in sizeof(*tab) * len + 1, I'll go with the idea both ends should be included.

对 len

Miscalculation of len

而不是循环，只需减去

//len = min;
//while (len < max)
//    len++;
len = max - min + 1;

分配计算错误

很好使用 sizeof * pointer ，但是+ 1没什么意义.如果有的话， ... * len +1 应该是 ... *(len +1).+1是通过上述解决方法处理的.在 C 中也不需要强制转换.

Good to use sizeof *pointer, yet the + 1 makes little sense. If anything the ... * len + 1 should have been ... * (len + 1). Yet the +1 is handled with the above fix. Also cast not needed in C.

// tab = (int *)malloc(sizeof(*tab) * len + 1);
tab = malloc(sizeof *tab * len);

错误分配

代码重复分配了相同的 * tab 位置.

Code repeatedly assigned the same *tab location.

//while (min < max)
//{
//    *tab = min;
//    min++;
//}
for (int i = min; i <= max; i++) {
  tab[i - min] = i;
}

没有分配错误检查，也没有最小，最大验证

No allocation error checking nor min, max validation

使用 mix-min

Potential for int overflow with mix - min

一定要免费分配

替代

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

int *ft_range(int min, int max) {
  if (min > max) {
    return NULL;
  }
  size_t len = (size_t)max - min + 1;

  int *tab = malloc(sizeof *tab * len);
  if (tab == NULL) {
    return NULL;
  }

  for (size_t i = 0; i < len; i++) {
    tab[i] = int(min + i);
  }

  return tab;
}

int main() {
    int mn = 0;
    int mx = 10;
    int *ft = ft_range(mn, mx);
    if (ft) {
      int *p = ft;
      for (int i = mn; i <= mx; i++) {
        printf("%d ", *p++);
      }
      free(ft);
    }
    return 0;
}

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