Using Realloc in C

问题:

Its really a post for some advice in terms of the use of realloc, more specifically, if I could make use of it to simplify my existing code. Essentially, what the below does, it dynamically allocate some memory, if i goes over 256, then the array needs to be increased in size, so I malloc a temp array, with 2x the size, memcpy etc. ( see below ).

I was just wondering if realloc could be used in the below code, to simplify it, any advice, sample code, or even hints on how to implement it is much appreciated!

Cheers.

void reverse(char *s) {
char p;

switch(toupper(s[0])) 
{
    case 'A': case 'E': case 'I': case 'O': case 'U':
        p = s[strlen(s)-1];
        while( p >= s )
            putchar( p-- );
        putchar( '\n' );
        break;
    default:
        printf("%s", s);
        break;
}
printf("\n");
    }

    int main(void) {
char c;
int buffer_size = 256;
char *buffer, *temp;
int i=0;

buffer = (char*)malloc(buffer_size);
while (c=getchar(), c!=' ' && c!='\n' && c !='\t') 
{
    buffer[i++] = c;
    if ( i >= buffer_size )
    {
        temp = (char*)malloc(buffer_size*2);
        memcpy( temp, buffer, buffer_size );
        free( buffer );
        buffer_size *= 2;
        buffer = temp;
    }
}
buffer[i] = '\0';
reverse(buffer);

return 0;

}

回答1:

Yes is the short answer. Here's how it would look:

if ( i >= buffer_size )
{
    temp = realloc(buffer, buffer_size*2);
    if (!temp)
        reportError();
    buffer_size *= 2;
    buffer = temp;
}

Note that you still need to use a temporary pointer to hold the result of realloc(); if the allocation fails you still have the original buffer pointer to the still-valid existing buffer.

回答2:

Realloc is pretty much exactly what you're looking for - you can replace that entire block inside the if ( i >= buffer_size ) with something like:

buffer = (char*)realloc(buffer, buffer_size*2);
buffer_size *= 2;

Notice that this ignores the error condition (if the return from realloc is NULL); catching this condition is left to the reader.

回答3:

Yes, realloc could be used to slightly simplify your code. If you're not interested in error-handling, then this:

char *tmp = malloc(size*2);
memcpy(temp, buffer, size);
free(buffer);
buffer = tmp;

is essentially equivalent to this:

buffer = realloc(buffer, size*2);

If you are interested in error-handling (and you probably should be), then you will need to check for NULL return values. This is true of your original code too.

回答4:

Yes, to simplify your code, you can replace

if ( i >= buffer_size )
{
    temp = (char*)malloc(buffer_size*2);
    memcpy( temp, buffer, buffer_size );
    free( buffer );
    buffer_size *= 2;
    buffer = temp;
}

with

if ( i >= buffer_size )
    buffer = realloc(buffer, buffer_size *= 2);

This does not take into account error checking, so you will need to check to make sure realloc doesn't return NULL.