Replace char in string with another string in C - c

Question

I'm trying to replace ' ' (space) with '___' (triple underscore) in C.
Here is my code:
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
int main()
{
char *a = "12 34 56";
int a_l = strlen(a);
printf("str1: \"%s\" (%d)\n", a, a_l);
char *b = "___";
int b_l = strlen(b);
printf("str2: \"%s\" (%d)\n", b, b_l);
for (int i = 0; i < a_l; i++) {
if (a[i] == ' ') {
char *o = malloc(a_l + b_l);
strncpy(o, a, i);
strncpy(o + i, b, a_l);
//strncpy help
printf("out: \"%s\"\n", o);
}
}
return 0;
}
I think that it is right so far, but I need to replace the comment line with correct strncpy (take the rest of string a (excluding space) and append it to string o). So the output should be like this:
str1: "12 34 56" (8)
str2: "___" (3)
out: "12___34 56"
out: "12 34___56"
If there are other mistakes in my code, please tell me.
UPD: This shouldn't replace all spaces in a loop. If the source string contains 8 spaces, there should be 8 lines printed and in each line only one space should be replaced.

Answer 1

You are overcomplicating this so much that I just TL;DR.
Some remarks that you might surely want to read, learn, embrace well and use:
I. int is not for string lengths and stuff. size_t is for string lengths and stuff.
II. String literals cannot be modified, so using the legacy char * type for assigning them to a variable is no good by any means, const-qualify that poor pointer base type.
III. Use VLAs instead of malloc() if dynamic memory management is not really needed (we're not living in 1989 anymore).
IV. NUL-terminate your strings because C stdlib routines expect you to do so.
int main()
{
const char *in = "foo bar baz";
int nspc = 0;
for (const char *p = strchr(in, ' '); p; p = strchr(p + 1, ' '))
nspc++;
char buf[strlen(in) + nspc * 2 + 1];
memset(buf, 0, sizeof(buf));
const char *s = in;
for (const char *p = strchr(s, ' '); p; p = strchr(s, ' ')) {
strncat(buf, s, p - s);
strcat(buf, "___");
s = p + 1;
}
const char *end = in + strlen(in);
strncat(buf, s, end - s);
printf("%s\n", buf);
return 0;
}

Answer 2

You can try this one. The problem comes from the fact that a_l + b_l in your malloc is always the same value. It doesn't affect by number of spaces.
int count = 0, index = 0;
for (int i = 0; i < a_l; ++i) {
if (a[i] == ' ') {
count++;
}
}
const char *o = malloc(a_l + 2 * count + 1); // 2 is because you add 3 new symbols, but remove 1 so 3 - 1 = 2
memset(o, 0, sizeof(o));
for (int i = 0; i < a_l; ++i) {
if (a[i] != ' ')
o[index++] = a[i];
else {
o[index++] = '_';
o[index++] = '_';
o[index++] = '_';
}
}

Answer 3

Without using library function!!!
while(*string)
{
if(*string=='\\')
{
*string++;
while(repl_len--)
*dest++ = *repl_string++;
}
else
{
*dest++ = *string++;
}
repl_string = temp_repl_string;
repl_len = temp_repl_len;
}

Answer 4

#include <string.h>
#include <stdlib.h>
#include <stdio.h>
size_t my_strlen(const char *str, size_t *spc){
size_t len;
*spc = len = 0;
while(*str){
++len;
if(*str++ == ' ')++*spc;
}
return len;
}
int main(void){
char *a = "12 34 56";
size_t spc;
int a_l = my_strlen(a, &spc);
printf("str1: \"%s\" (%d)\n", a, a_l);
char *b = "___";
int b_l = strlen(b);
printf("str2: \"%s\" (%d)\n", b, b_l);
char *p, *o = malloc(a_l - spc + spc * b_l + 1);
p=o;
for (int i = 0; i < a_l; i++) {
if(a[i] == ' ') {
strncpy(p, b, b_l);
p += b_l;
} else {
*p++ = a[i];
}
}
*p = '\0';
printf("out: \"%s\"\n", o);
free(o);
return 0;
}

Answer 5

I see that many answers have been added, but this may have been done very simply with a single loop. Since the string is very short, you can sacrifice memory over CPU and allocate an array 3 times +1 bigger than the original string, and be sure that it won't be overflowed:
const char* string= "12 34 56";
size_t length= strlen(string);
char result[length*3 +1];
unsigned int index= 0;
memset(result, '\0', length*3+1);
for(int i=0; i<length; i++)
{
if(string[i]!= ' ')
{
result[index++]= string[i];
}
else
{
strcat(result,"___");
index+= 3;
}
}

Answer 6

I found another thread and after reading the answers, I figured out that the right line should look like this:
strncpy(o + i + b_l, a + i + 1, a_l - i);
And after few fixes that were suggested in this thread my code looks like this:
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
int main()
{
char *a = "12 34 56";
size_t a_l = strlen(a);
printf("str1: \"%s\" (%d)\n", a, a_l);
char *b = "___";
size_t b_l = strlen(b);
printf("str2: \"%s\" (%d)\n", b, b_l);
for (int i = 0; i < a_l; i++) {
if (a[i] == ' ') {
char *o = malloc(a_l + b_l);
strncpy(o, a, i);
strcpy(o + i, b);
strncpy(o + i + b_l, a + i + 1, a_l - i);
printf("out: \"%s\"\n", o);
free(o);
}
}
return 0;
}
And this prints the desired putput.