merge two strings of different sizes in C - c

Question

How can I merge character by character two strings of different sizes to the n-th element of A and m-th element of B. I can easily do it when they have the same size (assuming here n=m). But cannot figure out how to handle this exception.
My working zip code is as follows:
char * zip(char *A, char *B, int n)
{
char *C;
int i;
C = malloc((2*n) * sizeof *A);
for(i=0; i<n; i++) {
C[(2*i)]=A[i];
C[(2*i)+1]=B[i];
}
return C;
}
But instead to pass just int n, I would like to pass also int m where n is to merge the n first elements of A and the m first elements of B. Hence passing the following input to the new_zip(char *A, char *B, int n, int m). A="rslxyzkw"; B="eutingxyz";n=3; m=6, I would get "resulting".

Answer 1

like this:
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
char *new_zip(char *A, char *B, int n, int m){
assert(A != NULL && B != NULL && n >= 0 && m >= 0);
char *C = malloc(n + m + 1);//+1 for NUL
if(!C){
perror("malloc:");
return NULL;
}
int i = 0;
while(n + m > 0){
if(n > 0 && *A){
C[i++] = *A++;
--n;
}
if(m > 0 && *B){
C[i++] = *B++;
--m;
}
}
C[i] = 0;
return C;
}
int main (void){
char *result = new_zip("rslxyzkw", "eutingxyz", 3, 6);
printf("'%s'\n", result);
free(result);
return 0;
}

Answer 2

Instead of looping until a certain length you can loop until you run out of characters. C strings have a null character at the end so after you allocate the memory you can loop as long as both aren't a null character. All you need to do then is only add non-null characters to your output string.
#include <stdio.h> /* printf */
#include <stdlib.h> /* malloc, free */
#include <string.h> /* strlen */
char *zip(char *a, char *b)
{
char *c = malloc( (strlen(a)+strlen(b)+1) * sizeof(char) ), *p = c;
if(c)
{
while(*a || *b) /* while either string has characters */
{
if(*a) *p++ = *a++; /* add a character from a if non-null */
if(*b) *p++ = *b++; /* add a character from b if non-null */
}
*p='\0'; /* finish the string with a null character */
}
return c;
}
int main()
{
char *a = "This is a string";
char *b = "This is another longer string";
char *c = zip(a,b);
if(c)
{
printf("zip(%s,%s) = %s\n",a,b,c);
free(c);
}
return 0;
}

Answer 3

The following could work.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char* zip(char* A, char* B) {
char *C;
int k = 0;
C = (char*)malloc(strlen(A)+strlen(B)+1);
while (*A != '\0' || *B != '\0') {
if (*A != '\0') {
C[k++] = *A;
++A;
}
if (*B != '\0') {
C[k++] = *B;
++B;
}
}
C[k] = '\0';
return C;
}
int main() {
char *A = "123456", *B = "abcd", *C;
C = zip(A, B);
puts(C);
return 0;
}

Answer 4

The below code would work in the following way:
First do alternate merging from strings s1 and s2 based on the minimum value of m and n.
The second part would take care of appending the remaining elements either from s1 or from s2.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
char* merge (char *s1, char *s2, int m, int n)
{
char *s = (char *) malloc(m + n + 1);
int min = (m < n)? m: n;
int i = 0, j = 0, k = 0;
int count = 0;
/* Alternate merge from s1 and s2 to s*/
while (count < 2 * min) {
if (count % 2 == 0) {
s[k++] = s1[i++];
} else {
s[k++] = s2[j++];
}
count++;
}
/* Append the remaining elements from s1 or s2 to s*/
if (m > min) {
for (count = 0; count < m - min; count++) {
s[k++] = s1[i++];
}
} else if (n > min) {
for (count = 0; count < n - min; count++) {
s[k++] = s2[j++];
}
}
s[k++] = '\0';
return s;
}
int main()
{
char *s1 = "rslxyzkw";
char *s2 = "eutingxyz";
char *s = merge(s1, s2, 3, 6);
printf ("%s\n", s);
}