convert characters in a string C

Question

I am working on a program to change take a word or multiple words from the user(at most 100 characters). for example if the user put in dakka dakka, they would get d@kk@ d@kk@ printed out. I am doing something wrong and it only prints the first word of what I type.

#include <stdio.h>
#include <string.h>

int main()
{
 char utext[100];
 int i, len;
 char c;

 len = strlen(utext);

 printf("type in your message to be converted: \n");
 fgets( utext, 100, stdin );

 for ( i = 0; i < len; ++i )
  {
  c = utext[i];

  if (c == 'a')
   c = '@';
 printf("%c", c);
  }

return 0;
}

Change fgets( utext, 100, stdin ); to if (NULL==fgets( utext, sizeof(utext), stdin)) return -1; — Weather Vane
– Weather Vane, Commented May 25, 2015 at 17:08

Iharob Al Asimi · Accepted Answer · 2015-05-25 16:58:21Z

You are calling strlen() on an uninitailized array.

The strlen() function searches for the terminating '\0' which is not in utext before you call fgets().

Also, you don't want to iterate over the 100 characters, which would be done if you change strlen() with sizeof(), because that will give you the size of the array in bytes, reason for which

fgets(utext, sizeof(utext), stdin);

is ok, and not just ok, it's better because now you can change the size of the array without needing to alter this line.

For the for loop, I would recommend using two facts

fgets() reads the trailing '\n' which is inserted by the user when Enter is pressed, and is almost mandatory.
Any valid c string, i.e. one that would return it's length when passed to strlen(), must have a terminating '\0', so if '\n' is not present for some reason, then '\0' would be.

From the two points above, the for loop should look like

for (i = 0 ; ((utext[i] != '\n') && (utext[i] != '\0')) ; ++i)
 {
    c = utext[i];
    if (c == 'a')
        c = '@';
    printf("%c", c);
 }

Bence Gedai · Accepted Answer · 2015-05-25 17:30:28Z

1

You might consider reading in by characters, then you dont even have to store the string. A code like this would work:

int c;
while((c = getchar()) != EOF)
{
    if(c == 'a')
        printf("@");
    else
        printf("%c", c);
}

edited May 25, 2015 at 17:30

answered May 25, 2015 at 17:14

Bence Gedai

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3 Comments

Weather Vane Over a year ago

getchar does not return NULL or a char. It returns an int the character value read, or EOF.

Bence Gedai Over a year ago

You're right, I edited the code. Although I tried it and it worked with char and NULL.

Weather Vane Over a year ago

while((c = getchar()) != NULL) gives the compiler warning warning C4047: '!=' : 'int' differs in levels of indirection from 'void *', also a char won't hold EOF without truncating it to 0xFF making it indistinguishable from the char 0xFF.

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