Modifying an array makes printf unusable [duplicate] - arrays

Question

This question already has answers here:
Why do I get a segmentation fault when writing to a "char *s" initialized with a string literal, but not "char s[]"?
(19 answers)
Closed 12 days ago.
I am trying to make an array of modifiable length, i defined a function called "add" that should add a char to the last of the array, but this result in putchar and printf not working. Why is this happening? and how can i fix it?
the output should be "hix", and the output is apparently ""
#include <stdio.h>
typedef struct
{
char* ptr;
size_t len;
}
bytes;
void add(bytes text, char chr)
{
text.ptr[text.len++] = chr;
}
bytes parse(char text[])
{
size_t index = 0;
while (text[index]) ++index;
return (bytes) {text, index};
}
void print(bytes text)
{
for (size_t index = 0; index < text.len; ++index)
{
putchar(text.ptr[index]);
}
}
int main()
{
bytes str = parse("hi");
add(str, 'x'); // if i remove this line "print" works, but only prints "hi"
print(str);
return 0;
}

Answer 1

When you try to modify string literal you will get memory overflow allocation problem.
So Your final code with allocation memory to sort the problem of Segmentation fault (core dumped)
#include <stdio.h>
#include <stdlib.h>
typedef struct {
char * ptr;
size_t len;
}
bytes;
void add(bytes * text, char chr) {
text -> ptr[text -> len++] = chr;
}
bytes parse(char text[]) {
size_t index = 0;
char * new_ptr = NULL;
while (text[index]) ++index;
new_ptr = (char * ) malloc(index + 1);
if (new_ptr == NULL) {
printf("Memory allocation failed!\n");
exit(1);
}
for (int i = 0; i < index; i++) {
new_ptr[i] = text[i];
}
new_ptr[index] = '\0';
return (bytes) {
new_ptr,
index
};
}
void print(bytes text) {
for (size_t index = 0; index < text.len; ++index) {
putchar(text.ptr[index]);
}
}
int main() {
bytes str = parse("hi");
add( & str, 'x');
print(str);
free(str.ptr);
return 0;
}
Result :
hix
You can check by compile this code on C compiler