C语言中基于用户输入的动态增长数组

时间:2019-05-22 09:56:07

标签: c dynamic-arrays

我需要创建一个程序,该程序:

  
      
  1. 最初分配一个数组以读取并保持5个温度。
  2.   
  3. 提示用户输入温度并在输入温度时输入值-100.0
  4.   
  5. 如果用户填写了数组,则程序应   动态分配一个新数组,其大小是原来的两倍。
  6.   
  7. 将旧值复制到新数组中。   取消分配旧数组。
  8.   
  9. 继续读取新数组。
  10.   
  11. 完成后打印出新数组
  12.   

我对C完全陌生,有点卡住了。我知道如何创建动态数组,但是我不知道如何创建一个新数组,一旦旧数组被填满,该数组就会不断增长。

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

int main(void){
    int i,k; //loop count
    int j = 5; //initial array size
    int* temp = malloc(sizeof(int)*j); 
    int* newtemp;

    for (i = 0; i < j; i++){ //loop to read in temperature
        printf("enter temperature: ");
        scanf("%d",(temp+i));
        if (i=j){
        j = j*2; //double the size of initial array
        int* newtemp = malloc(sizeof(int)*j);
        strcpy(*newtemp,temp); // copy string
        for (k = 0; k < j; k++){ //loop to read in temperature
            printf("enter temperature: ");
            scanf("%d",(temp+i+k));
            }
        }
        switch (temp[i]){
            case (-100):
            temp[i] = '\0';
            i = 5; //loop ends
            break;
        }    
    }
    return 0;
}

错误消息:

tempp.c:18:16: warning: passing argument 1 of ‘strcpy’ makes pointer from integer without a cast [-Wint-conversion]
         strcpy(*newtemp,temp);
                ^
In file included from tempp.c:3:0:
/usr/include/string.h:121:14: note: expected ‘char * restrict’ but argument is of type ‘int’
 extern char *strcpy (char *__restrict __dest, const char *__restrict __src)
              ^~~~~~
tempp.c:18:25: warning: passing argument 2 of ‘strcpy’ from incompatible pointer type [-Wincompatible-pointer-types]
         strcpy(*newtemp,temp);
                         ^~~~
In file included from tempp.c:3:0:
/usr/include/string.h:121:14: note: expected ‘const char * restrict’ but argument is of type ‘int *’
 extern char *strcpy (char *__restrict __dest, const char *__restrict __src)

我知道我的代码很乱,而且我真的不知道在不断增长的同时重新分配新数组的正确方法。 请帮我解决一下这个。谢谢!

4 个答案:

答案 0 :(得分:2)

如何使用realloc工具呢?

void printArray(double *array, int size){
    for(int i=0; i<size; i++){
        printf("%.1lf ", array[i]);
    }
    putchar('\n');
}

int main(void){
    int size = 5;
    double *array = malloc(size * sizeof(double)); 
    double temperature;
    int i = 0;

    while(1){
        if(temperature == -100.0)
            break;
        if(i == size){
            size *= 2;
            array = realloc(array, size * sizeof(double));
        }
        scanf("%lf", &temperature);
        array[i] = temperature;
        printArray(array, size);
        i++;
    }
    free(array);
    return 0;
}

答案 1 :(得分:0)

#define INITIALSIZE 5

typedef struct 
{
    size_t size;
    size_t index;
    int data[];
}DATA_t;

DATA_t *addData(DATA_t *data, int val)
{
    if(!data)
    {
        data = malloc(INITIALSIZE * sizeof(data -> data[0]) + sizeof(*data));
        /* malloc result checks */
        data -> size = 0;
        data -> index = 0;
    }
    if((data -> index + 1) == data -> size)
    {
        size_t newsize = data -> size * 2 ;
        DATA_t *newdata = malloc(newsize * sizeof(data -> data[0]) + sizeof(*data));
        /* malloc result checks */
        memcpy(newdata, data, data -> size * sizeof(data -> data[0]) + sizeof(*data));
        newdata -> size = newsize;
        free(data);
        data = newdata;
    }
    data -> data[data -> index++] = val;
    return data;
}

用法:

DATA_t *mydata = NULL;

while(condition)
{
    mydata = addData(mydata, ReadValue());
    /* ----- */
}

答案 2 :(得分:0)

您可以尝试声明两个数组,并在它们之间切换,就像这样:

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

int main(void){

    int currentArraySize = 5;

    int* temperatures1 = malloc(sizeof(int)*currentArraySize);
    int* temperatures2 = NULL;
    int temperaturesSlot = 0;
    int temperature = 0;
    int index = 0;


    while(1){
        if (index == currentArraySize){
            switch (temperaturesSlot){
                case 0:
                    temperatures2 = malloc(sizeof(int)* 2 *currentArraySize);
                    memcpy(temperatures2, temperatures1, currentArraySize * sizeof(int));
                    free(temperatures1);
                    temperatures1 = NULL;
                    temperaturesSlot = 1;
                    break;
                case 1:
                    temperatures1 = malloc(sizeof(int)* 2 *currentArraySize);
                    memcpy(temperatures1, temperatures2, currentArraySize * sizeof(int));
                    free(temperatures2);
                    temperatures2 = NULL;
                    temperaturesSlot = 0;
                    break;
            }
            currentArraySize *= 2;
        }

        printf("enter temperature: ");
        scanf("%d",(&temperature));
        if (temperature == -100){
            break;
        }
        else if (temperaturesSlot == 0){
            temperatures1[index] = temperature;
        }
        else{
            temperatures2[index] = temperature;
        }
        ++index;
    }

    for (int i = 0; i < index; ++i){
        if (temperaturesSlot == 0){
            printf("%d, ", temperatures1[i]);
        }
        else{
            printf("%d, ", temperatures2[i]);
        }
    }
}

答案 3 :(得分:-2)

如果您是初学者,有一个简单的解决方案。假设输入已经在第一个数组myArray中。

//both arrays can be created dynamically, but are not in this instance for simplicity of the example.
int myArray[5] = { 19,10,11,5,7 };    //initial array of size 5
int doubleMyArray[10] = { 0 };        //array of size double, filled with 0s
int i;

//size of an array: sizeof(myArray)/sizeof(myArray[0])
for (i = 0; i < (sizeof(myArray) / sizeof(myArray[0])); i++) {
    doubleMyArray[i] = myArray[i];
}

然后,您可以将新元素(如果有)放入doubleMyArray [i]中,将每个输入(i ++)的i值加1,直到i <[array size]或输入中的用户输入为-100环。

您的阵列并没有持续增长。首先是由5个元素组成的数组,然后是由10个元素组成的新数组。将所有元素移动到第二个数组后,您将从第一个数组中分配内存,并在最后输出新数组,以检查期望的元素是否存在。

此外,没有关于动态分配的信息。如果您创建了一个函数以将前5个输入带入第一个数组(在该函数内本地创建),然后将第二个数组从该函数返回到main()中,且第一个数组的元素已在其中,则第一个数组将自动取消分配自身,仅在main()中留下第二个数组,因为第一个数组仅存在于该函数的范围内。