How to copy with memcpy? - c

Question

char *concat(char *num1, const char *num2, int index) {
int length1 = strlen(num1);
int length2 = strlen(num2);
int lengthNum = 0;
char *num = malloc(length1 + length2 + 1);
if (num == NULL) {
free(num);
return NULL;
}
// memcpy(num, num1, length1);
// memcpy(num + length1, num + index, length2 + 1);
for (int i = 0; i < length1; i++) {
num[lengthNum] = num1[i];
lengthNum++;
}
for (int i = index; i < length2; i++) {
num[lengthNum] = num2[i];
lengthNum++;
}
return num;
}
I tried to use memcpy, but than my program doesn't work correctly (copies wrongly, but valgrind doesn't show an error).
But when I use two for loops instead, it works properly, but than valgrind shows an error
uninitialised value was created by a heap allocation.
How to use properly memcpy in this case?

Answer 1

memcpy(num, num1, length1);
memcpy(num + length1, num2, length2 + 1);

Answer 2

Your program has multiple issues:
freeing a null pointer when malloc failed is useless (but harmless).
you allocate length1 + length2 + 1 bytes, which is most likely too large as you intend to copy from index in the second string.
you should use type size_t for the lengths and offsets.
you copy from num2 + index without checking that index is a valid offset inside the string num2.
you should set a null terminator at the end of the new string.
Here is a modified version:
#include <stdlib.h>
#include <string.h>
char *concat(const char *num1, const char *num2, size_t index) {
size_t length1 = strlen(num1);
/* skip index bytes in num2 */
while (index --> 0 && *num2)
num2++;
size_t length2 = strlen(num2);
char *num = malloc(length1 + length2 + 1);
if (num != NULL) {
size_t j = 0;
while (*num1) {
num[j++] = *num1++;
}
while (*num2) {
num[j++] = *num2++;
}
num[j] = '\0';
}
return num;
}
and using memcpy:
#include <stdlib.h>
#include <string.h>
char *concat(const char *num1, const char *num2, size_t index) {
size_t length1 = strlen(num1);
/* skip index bytes in num2 */
while (index --> 0 && *num2)
num2++;
size_t length2 = strlen(num2);
char *num = malloc(length1 + length2 + 1);
if (num != NULL) {
memcpy(num, num1, length1);
memcpy(num + length1, num2, length2 + 1);
}
return num;
}
concat returns a pointer to an allocation array. It is the responsibilty of the caller to free this object after use. For example:
#include <stdio.h>
int main() {
char *p = concat("Hello", "dear world", 4);
if (p != NULL) {
printf("%s\n", p);
free(p);
}
return 0;
}