Dynamically allocating text, and error in free()

Question

I’m trying to allocate dynamically, but the error is appearing:

Error in . /name. c; invalid : 0x000001c18834

My code:

// FUNCAO // 
char *nome(text[])  
{    
    int n3 = 0;  
    int n2 = 0;     
    char *n = NULL;  

    while((n2 = getchar()) != '\0' && n2 != EOF)  
    {  
        n3++;  
        n = realloc(n, n3*sizeof(char));  
        if(n == NULL)  
        {  
            puts("Erro ao realocar!");  
            exit(0);  
            free(n);  
        }   
        *(n+(n3-1)) = n2;   
    }  
    *(n+n3) = '\0';  
    return n;
}

// USANDO //   
int main(void)  
{  
    char *name = nome("Nome:");  
    while(*name != '\0')  
    printf("%c", *name++);    
    free(name); // aqui ta o problema, sem o free roda normal agora com o free me da esse erro
}        

Answer 1

The code has some problems, even is not readable. I simplified a little. The most serious problem is that it keeps in name the pointer to the allocated location. It cannot be modified so that the free at bandstand. When increment will try to free memory instead that there was no allocation and generates the error. The correct is to create an independent iteration variable and increment it, so do not move me name.

I didn’t see if you had any other problems.

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

char *nome(char text[]) {    
    int n3 = 0;
    int n2 = 0;
    char *n = NULL;
    while((n2 = getchar()) != '\0' && n2 != EOF) {
        n = realloc(n, ++n3);
        if (n == NULL) {
            puts("Erro ao realocar!");
            exit(0);
        }
        *(n + (n3 - 1)) = n2;   
    }
    *(n + n3) = '\0';  
    return n;
}   

int main(void) {  
    char *name = nome("Nome:");
    char *iterador = name;
    while(*iterador != '\0') printf("%c", *iterador++);
    printf("%s", name); //bem mais simples, certo?
    free(name);
}

Behold working in the ideone. And in the repl it.. Also put on the Github for future reference.