Replacing a character in a string (char array) with multiple characters in C

Question

C is not a managed language, and it's been a while since I have used an unmanaged language. I need to create a block of code which searches through a character array, and replaces all instances of '2' with 'Ba'. This is quite simple, except for the fact that now the resulting string will be larger than the original, so I need to account for this somehow, and I don't remember how to do this.

Here is try number 4, using pointers:

//Move next 150 characters after replacing all 2 with Ba:
    numberread = read(XYZFile, datatoreadWWW, sizeof(datatoreadWWW));
    int y = strlen(datatoreadWWW); //should be 150.
    int additionallength = y;
    i = 0;
    while(i < y) {
      if(datatoreadWWW[i] == '2') {
        additionallength++; //Since we need to replace any '2' with 'Ba', we should add one character for the a.
      }
      i++;
    }
    //And now use additionallength to create the new char array to be placed in WWW, called newstring:
    char newstring[additionallength];
    int j = 0;
    const char *in = datatoreadWWW;
    char *out = newstring;
    while(*in) {
      if (*in == '2') {
    *out++ = 'B';
    *out++ = 'a';
      }
      else {
    *out++ = *in;
      }
      in++;
    }
    *out++ = '\0';

And I'm still confused/stuck on this, I'm getting garbage values written into the WWW file such as "^Q" and such. Not sure what that means, but it doesn't seem correct.

Edit: The program above seems to be working now. The problem was, if anyone ever reads this post as a resource, that I was adding +1 to additionallength, but this was happening in every iteration, so every so many characters there was an extra empty, null space being put in to the file, which was incorrect. Regardless, I think we've learned that is important to review pointers whenever working in C, as this is the easiest way to accomplish this.

"managed" or "unmanaged" has nothing to do with this. I think you are referring to the lack of a standard container with an insert operation. — M.M
– M.M, Commented Sep 9, 2014 at 2:16
'Ba' is a multi-byte character constant , it is still a single character after all that. What you actually want is two characters, 'B' and 'a'. As you say, you need to have enough space because the result will be larger than the original. Try writing some code and then posting it. — M.M
– M.M, Commented Sep 9, 2014 at 2:18
C is a managed language, it's just the developer (rather than the runtime) doing the management :-) — paxdiablo
– paxdiablo, Commented Sep 9, 2014 at 2:24

Sergey Kalinichenko · Accepted Answer · 2014-09-09 02:26:20Z

2

You can do it like this:

Compute the length of the new array by adding the number of '2's to strlen of your string plus one for null terminator
Allocate the required number of chars using malloc
Go through your string in a loop; if you see '2', add 'B' followed by 'a' to the output;
Otherwise, copy the current character to the output.

Note: now that your string is allocated dynamically, you need to call free once you are done using it.

const char* str = ...
int cnt = strlen(str)+1;
for (const char *p = str ; *p ; cnt += (*p++ == '2'))
    ;
char *res = malloc(cnt);
char *out = res;
const char *in = str;
while (*in) {
    if (*in == '2') {
        *out++ = 'B';
        *out++ = 'a';
    } else {
        *out++ = *in;
    }
    in++;
}
*out++ = '\0';

Demo.

edited Sep 9, 2014 at 2:26

answered Sep 9, 2014 at 2:19

Sergey Kalinichenko

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7 Comments

5gon12eder Over a year ago

Or, if you don't want to loop over the string twice, you could optimistically pre-allocate a buffer that is only “somewhat” larger than the input and realloc()ate it if it turns out that it was too short.

Musicode Over a year ago

Can this be accomplished without making any allocation calls? I simply created a new character array with size of the original, plus the amount of 2s that were present, plus 1 for the null terminator.

Sergey Kalinichenko Over a year ago

@Musicode That would work, too, as long as the result is used within the same function. Replace the malloc line with char res[cnt];, and the code would continue to work.

Sergey Kalinichenko Over a year ago

@Musicode You forgot the last line (adding null terminator).

Sergey Kalinichenko Over a year ago

@Musicode Your last attempt is incorrect in other ways, too: you try to use a single index j instead of two pointers in and out in my example. That is why your code skips over some characters, leaving gaps. If you do not like pointers, use indexes; it's all the same. Once the loop is over, put '\0' in the character at your writing index.

|

Jim Balter · Accepted Answer · 2014-09-09 06:38:14Z

1

Many of the solutions are excessively complex, pointlessly verbose, and are inefficient (e.g., strlen means three scans when only two are needed).

int newlen = 1;

for (char* p = str; *p; p++)
    newlen += *p == '2' ? 2 : 1;

char newstr[newlen];

for (char *p = str, *q = newstr;; p++)
    if (*p == '2')
    {
        *q++ = 'B';
        *q++ = 'a';
    }
    else if (!(*q++ = *p))
        break;

edited Sep 9, 2014 at 6:38

answered Sep 9, 2014 at 3:20

Jim Balter

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3 Comments

Musicode Over a year ago

I tried this method but I am now getting garbage values in my WWW file that I am writing to...

John Hascall Over a year ago

char newstr[newlen]; is not legal C, use char * newstr = malloc(newlen);

Jim Balter Over a year ago

@Musicode Perhaps you're writing it incorrectly, or perhaps your input is garbled.. I notice that you're using read ... that's not a good idea and doesn't NUL-terminate the input. Use stdio.

nem035 · Accepted Answer · 2014-09-09 02:22:55Z

0

Here is an idea how to do this:

count the number of '2's in your char array and store in a variable, let's say count.
create a new char array, let's say new_word, with the size of your previous char array plus the number of 2's stored in the count variable.
iterate from the end of your input word, storing the characters in the new_word. Each time you reach a '2', you insert 'a', then 'B', and of course, update the indexes properly.

answered Sep 9, 2014 at 2:22

nem035

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Comments

Anbu.Sankar · Accepted Answer · 2014-09-09 03:07:13Z

0

You can try like this also..

main()
{
char a[]="abcabcabcabcabcabcabc";
char *p;
int i=0,c=0;

for(i=0;a[i];i++)
if(a[i]=='c')
c++;

printf("c=%d\n",c);
p=calloc(sizeof(a)+c,1);

for(i=0;a[i];i++)
{
static int j=0;
if(a[i]=='c')
    {
    *(p+j++)='n';
    *(p+j++)='s';
    }
 else
    {
    *(p+j++)=a[i];
    }
   }
   printf("%s\n",p);
  }

answered Sep 9, 2014 at 3:07

Anbu.Sankar

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