我想咨询关于realloc
的问题。
以下代码可以正常工作(没有警告):
#include <stdio.h>
#include <stdlib.h>
int main ()
{
int num=10;
int *vet;
int i;
for (i=0; i<num; i++)
{
/* allocate memory of vet to contains (i+1) int */
vet = (int*) realloc ( vet, (i+1) * sizeof(int) );
/* write numbers in the allocated memory */
vet[i] = 321 + i;
}
/* print test, if all works I must see:
| 321 | 322 | 323 | 324 | 325 | 326 | 327 | 328 | 329 | 330 | */
printf ("| ");
for (i=0; i<num; i++)
printf ("%d | ", vet[i]);
printf ("\n");
return 0;
}
但是带有函数的相同程序却无法运行!编译器返回以下警告:
In function ‘main’:
14:10: warning: ‘vet’ is used uninitialized in this function [-Wuninitialized]
代码如下:
#include <stdio.h>
#include <stdlib.h>
void memoria (int *, int);
int main ()
{
int *vet, num=10;
memoria (vet, num);
/* print test, if all works I must see:
| 321 | 322 | 323 | 324 | 325 | 326 | 327 | 328 | 329 | 330 | */
int i;
printf ("| ");
for (i=0; i<num; i++)
printf ("%d | ", vet[i]);
printf ("\n");
return 0;
}
void memoria (int *vet, int num)
{
int i;
for (i=0; i<num; i++)
{
/* allocate memory of vet to contains (i+1) int */
vet = (int*) realloc ( vet, (i+1) * sizeof(int) );
/* write numbers in the allocated memory */
vet[i] = 321 + i;
}
}
有人能告诉我为什么吗?非常感谢!
哦,而且在主函数中使用“随机”malloc的相同代码可以正常工作(与该函数一起)...
#include <stdio.h>
#include <stdlib.h>
void memoria (int *, int);
int main ()
{
int *vet, num=10;
/* ADDED MALLOC */
vet = (int*) malloc (1);
memoria (vet, num);
/* print test, if all works I must see:
| 321 | 322 | 323 | 324 | 325 | 326 | 327 | 328 | 329 | 330 | */
int i;
printf ("| ");
for (i=0; i<num; i++)
printf ("%d | ", vet[i]);
printf ("\n");
return 0;
}
void memoria (int *vet, int num)
{
int i;
for (i=0; i<num; i++)
{
/* allocate memory of vet to contains (i+1) int */
vet = (int*) realloc ( vet, (i+1) * sizeof(int) );
/* write numbers in the allocated memory */
vet[i] = 321 + i;
}
}
memoria(&vet,num);
。此外,不要将malloc或realloc的结果强制转换,因为它可能会使有价值的诊断信息无声,而且看起来很糟糕。 - Brandingcc版本4.8.1
,没有任何警告(已测试)! - ᴜsᴇʀ-O3
,但您需要在命令行上使用某种形式的-O
。 - Jonathan Leffler