Using realloc inside a function [duplicate]

问题:

This question already has an answer here:

• C Programming: malloc() inside another function 8 answers

My apologies, I know many related questions have already been asked, so I will keep it very simple.

Despite some years of programming I cannot find the correct syntax for resizing and modifying an array (or several) inside a function. For example, say I want a function to fill an array with a set of "n" numbers, where "n" is defined within the array:

int main(int argc, char *argv[]) {
    float *data = NULL
    int n = myfunction(data);
    for(i=0;i<n;i++) printf("%f\n",data[i]);
    free(data);
}

int myfunction(float *input) {
    int i,n=10;
    input = (float *) realloc( input, n*sizeof(float) );
    if(input!=NULL) {
        for(i=0;i<n;i++) input[i] = (float)i;
        return(n);
    else return(-1)
}

I know this will not work, as I probably need to use a pointer to a pointer, but I cannot resolve which combination of pointers, pointers-to-pointers, and address notation to use inside and outside the function to use.

Any simple suggestions appreciated!

回答1:

You need to pass a pointer to a pointer to myFunction

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

int myfunction(float **input) {
    int i,n=10;
    *input = realloc( *input, n*sizeof(float) );
    if(*input!=NULL) {
        for(i=0;i<n;i++) (*input)[i] = (float)i;
        return(n);
    }
    else return(-1);
}

int main(int argc, char *argv[]) {
    float *data = NULL;
    int n = myfunction(&data);
    int i;
    for(i=0;i<n;i++) printf("%f\n",data[i]);
    free(data);
    return 0;
}

回答2:

It's easiest to pass the old pointer to myfunction(), and have it return the new pointer (which might be the same as the old, if realloc() managed to grow the area in-place).

Note that realloc() can fail, in that case you don't want to lose track of the old memory which is still allocated so overwriting the same pointer without checking is a bad idea.