character array assignment error in C

Question

I have some code that is meant to read the first 3 characters from a character array that is read from a file, it was working then without changing anything is stopped working. The 'command' char array use to hold "and" but now often holds "add▒" and sometimes "and0" but I've only declared it to be 3 long yet it still manages to hold more. Is there some context I am missing?

//ORIGINALLY THIS WORKED

    for (i = 0; i < 3; i++){
        command[i] = line[i];
    }

    /*Interpret AND or ADD or JMP */
    if (strcmp(command,"and") == 0){
        hexLine[0] = changeHex(5);
    }else if (strcmp(command,"add") == 0){
            hexLine[0] = changeHex(1);
    }else if (strcmp(command,"jmp") == 0){
            hexLine[0] = changeHex(12);
    }
    printf("%s", command);

//AND NOW THIS DOESNT WORK

for (i = 0; i < 3; i++){
    command[i] = line[i];
}

/*Interpret AND or ADD or JMP */

if (strcmp(command,"and") == 0){
    hexLine[0] = changeHex(5);
}else if (strcmp(command,"add") == 0){
    hexLine[0] = changeHex(1);
}else if (strcmp(command,"jmp") == 0){
    hexLine[0] = changeHex(12);
}else if (strcmp(command,"ld ") == 0){
    hexLine[0] = changeHex(2);
}
printf("%s", command);

What exactly makes you think there's a fourth character in command? — aschepler
– aschepler, Commented Jun 1, 2014 at 22:06
Your for loop copies 3 characters but doesn't zero-terminate command. So strcmp won't behave the way you want. Put command[i] = 0; after your for loop. — lurker
– lurker, Commented Jun 1, 2014 at 22:12
Don't forget the null-character. See cplusplus.com/reference/cstring/strcmp — Klaus
– Klaus, Commented Jun 1, 2014 at 22:13
If your question has been answered, please don't edit your question to reflect that fact. Instead accept the answer you think is correct. — Nik Bougalis
– Nik Bougalis, Commented Jun 1, 2014 at 22:34

lurker · Accepted Answer · 2014-06-02 10:00:05Z

1

Your for loop copies 3 characters but doesn't zero-terminate command. So strcmp won't behave the way you want. Put command[i] = 0; after your for loop.

for (i = 0; i < 3; i++){
    command[i] = line[i];
}

command[i] = `\0';

As @Klaus points out in his comment: the above for loop assumes that you always have 3 valid characters to copy over. And, of course, command must be an array of at least 4 characters.

edited Jun 2, 2014 at 10:00

answered Jun 1, 2014 at 22:14

lurker

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5 Comments

user3697681 Over a year ago

Thank you so much, this fixed the error even with only 3 spaces.

user3629249 Over a year ago

stepping on a 4th entry in command when command is only defined as three chars will corrupt the stack. If you going to use this method, expand the size of command to 4 chars (still only read 3 chars.)

lurker Over a year ago

@user3629249 of course. Op's originsl code doesn't show the declaration of command but I have edited to make that clear in my response.

user3629249 Over a year ago

Generally, you should modify the code to accommodate the data rather than hacking data representation to accommodate the code.

lurker Over a year ago

@user3629249 indeed. But given that there is little information available for the OP's context, it's unclear whether the data is actually being "hacked" in this case.

user235244 · Accepted Answer · 2014-06-01 22:21:26Z

0

you need to put terminator character \0 in the end of the command char array

for (i = 0; i < 3; i++){
command[i] = line[i];
}

command[i + 1] = '\0';

answered Jun 1, 2014 at 22:21

user235244

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