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動的メモリmallocの使い方

Last updated at Posted at 2020-02-29

malloc

  • みんなの敵malloc
  • わからなくなりがちだからまとめ
  • コロナと一緒に倒しちゃいましょう!
  • #include < stdlib.h> です!
  • 0で初期化したい場合はcallocを使う

基本形

ポインタ = (型*)malloc(sizeof(型) * 大きさ)

例えば、

int *p = (int*)malloc(sizeof(int)*100);
free (p);

#include <stdio.h>
#include <stdlib.h>
int main(){
  int arraySize = 5;
  int* p;
  p = (int*)malloc(sizeof(int)*arraySize);
  
  int i;
  for(int i=0; i<arraySize; i++) p[i] = 2*i;
  for(int i=0; i<arraySize; i++) printf("p[%d] = %d\n", i, p[i]);

  free(p);
  return 0;
}

出力は
p[0] = 0
p[1] = 2
p[2] = 4
p[3] = 6
p[4] = 8

配列のコピー

  • #include < string.h>が必要です

memcpy(dest, src, size)

int a[] = {1, 2, 3, 4, 5};
int *b = (int*)malloc(sizeof(int)*5);
memcpy(b, a, sizeof(int) * 5);

malloc()関数

使い方の例

int *ptr = (int*)malloc(sizeof(int)*10);

if(ptr == NULL) exit(1); //メモリ確保に失敗した時はNULL

ptr[0] = 123;
ptr[1] = 555;

free(ptr);

構造体で使う

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

// point構造体の定義は省略

struct point *init_point(int x, int y){
  struct point *ptr = malloc(sizeof(struct point));

  ptr->x = x;
  ptr->y = y;

  return ptr;
}

// または

struct point *init_point(int x, int y){
  struct point *ptr = malloc(sizeof(struct point));

  *ptr = (struct point){.x = x, .y = y};
  return ptr;
}

int main(){
  struct point *ptr = init_point(10, 20);

  printf("%d\n", ptr->x);
  printf("%d\n", ptr->y);
}

あとがき

コロナのせいで予定なくなって暇になって書きました
超低クオリティですみません!!!!

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