replacing spaces with %20 in c program - c

Question

I have written the following program to replace spaces with %20.It works fine.
But it prints some garbage values for the pointer variable ptr though it might have been limited to 8 characters as the malloc assigns it 8 bytes of memory.
Can anyone tell me where did I go wrong here ?
Or is there any in place algorithm ?
void replaceSpaces(char *inputStr )
{
char *ptr;
int i,length, spaceCount=0;
int newLength,j;
for (length=0; *(inputStr+length)!='\0';length++ )
{
if (*(inputStr+length)==' ')
{
spaceCount++;
}
}
newLength = length + 2*spaceCount;
ptr = (char *)malloc(newLength*sizeof(char));
for ( i = length-1; i >=0; i--)
{
if (*(inputStr+i)==' ')
{
*(ptr+newLength-1)='0';
*(ptr+ newLength-2)='2';
*(ptr+newLength-3)='%';
newLength = newLength -3;
}
else
{
*(ptr+newLength-1) = *(inputStr+i);
newLength = newLength -1;
}
}
for ( i = 0; *(ptr+i)!='\0'; i++)
{
printf("%c",*(ptr+i));
}
}

Answer 1

Either use calloc() to allocate memory for ptr or terminate it with '\0' after allocation.
With your code, ptr never gets terminated with '\0'.
So either change
ptr = (char *)malloc(newLength*sizeof(char));
to
ptr = calloc(newLength*sizeof(char), sizeof(char));
Or add below line after allocating the ptr.
ptr[newLength] = '\0';

Answer 2

If you don't need the converted string, then you don't need to worry about conversion: you can just output characters directly.
void replace_spaces(const char *in) {
for (const char *p=in; *p; p++) {
if (*p == ' ') {
puts("%20");
} else {
putch(*p);
}
}
}
If you do need the converted string, then a useful pattern for this sort of code is to do the string conversion twice; once in "dry-run" mode where you do the conversion but don't write to the result, and once in "live" mode where you're actually writing. In the first pass, you calculate the needed length. This avoids duplication of logic, and makes it more obvious if you've counted the resulting length correctly. Here's some code in that style, with some test-cases.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char *replace_spaces(const char *in) {
char *result = 0;
for (int write = 0; write <= 1; write++) {
int n = 0;
for (const char *from = in; ;from++) {
if (*from == ' ') {
if (write) memcpy(result + n, "%20", 3);
n += 3;
} else {
if (write) result[n] = *from;
n++;
}
if (!*from) break;
}
if (!write) result = malloc(n);
}
return result;
}
int main(int argc, char**argv) {
const char*cases[] = {"Hello world", "abc", " sp sp sp "};
for (int i = 0; i < sizeof(cases) / sizeof(cases[0]); i++) {
char *rep = replace_spaces(cases[i]);
printf("'%s' -> '%s'\n", cases[i], rep);
free(rep);
}
return 0;
}

Answer 3

Assuming that this is going to work on fairly short strings, it is a bit silly to go through the strings twice. If, like in most cases, CPU is more valuable than memory, do:
char *dst = malloc(3*strlen(src) + 1);
char *result = dst;
while (*src) {
if (*src == ' ') {
strcpy(dst, "%20"); dst+=3;
}
else *dst++ = *src;
src++;
}
*dst = 0;
// result is result