Weird Symbols printed at end of string - C

Question

So, I've got a program that parses expressions in a line of text such as

11110000 & 11001100 ;

and evaluates the binary results. My code is parsing correctly and evaluating correctly, but for two of my test inputs (including the one above) my printf is also printing these weird symbols after each run.

eos$ ./interpreter < program02.txt
11000000 +
eos$ ./interpreter < program02.txt
11000000 2ñ
eos$ ./interpreter < program02.txt
11000000 "]
eos$ ./interpreter < program02.txt
11000000 ÒØ
eos$ ./interpreter < program02.txt
11000000 Ê
eos$ ./interpreter < program02.txt
11000000 òJ

The string is malloc'd like this

char *str = ( char * ) malloc ( ( getLength( src ) + 1 ) * sizeof( char ) );

And here is how the string is printed

char *str = binaryToString( val );
printf( "%s\n", str );

Any help would be awesome! Thanks!

probably forgot a null terminator, allowing printf() to run "past the end" of the string. — Marc B, Apr 22 '15 at 16:42
This `char *str = ( char * ) malloc ( ( getLength( src ) + 1 ) * sizeof( char ) );` is unecessarily ugly. -> `char *str = malloc(getLength(src) + 1);` — Iharob Al Asimi, Apr 22 '15 at 16:42
And I am sure that you forgot the `nul` terminator, there is no other possible reason for such behavior. When you post a question, you must post code taht reproduces the problem, otherwise it's not possible to find out what is wrong, you didn't post the very important `binaryToString()`. Also, I assume that `string.h` is forbiden, because there exists `strlen()`. — Iharob Al Asimi, Apr 22 '15 at 16:43
You're all correct, I forgot to add the null terminator, thanks so much! I feel stupid. — Eric Davidson, Apr 22 '15 at 16:54

Arite · Accepted Answer · 2015-04-22T17:25:47.527

Strings are null terminated in C. When you malloc() memory, it will be filled with whatever was in the block previously.

~~One solution is to fill the buffer with the null character \0 via memset() (found in string.h) after using malloc() like so:~~

int strLen = getLength(src) + 1;
char *str = (char*)malloc(strLen * sizeof(char));
memset(str, '\0', strLen); // Fill with null chars

Equally you could just write a \0 after the final character.

EDIT: This is not good advice according to iharob's comment. Taking this into account, and given you know the length of the string:

int strLen = getLength(src) + 1;
char *str = calloc(strLen, sizeof(char)); // Allocate strLen * sizeof(char)
if (str == NULL) {
    // Error allocating str - handle error
}
str[strLen - 1] = '\0'; // zero-based, so char is after final character

Is a better solution.

This is very bad advice in so many ways, 1. [Do not cast the result of `malloc()`](http://stackoverflow.com/a/605858/1983495), 2. `sizeof(char) == 1` **by definition**. 3. Your `memset()` is completely unnecessary, if you are allocating `strLen` characters, just `str[strLen - 1] = '\0'` is enough, using `memset()` for that might, hide bugs, when you try for example to debug the program with tools like [valgrind](http://www.valgrind.org). 4. There exists `calloc()`. 5. Don't dereference the pointer if you didn't check for `NULL` yet. — Iharob Al Asimi, Apr 22 '15 at 17:04

Weird Symbols printed at end of string - C

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