Weird Error: Abort trap while handling character array in C

Question

I want to store the binary value of each character in a string and store it in an array. But when i start messing around with functions like memset, i have no control over the debugging.

#include <stdio.h>
#include <string.h>
int main()
{
char str[8];
char *ptr = "Hello";
int i;

for(; *ptr != 0; ++ptr)
{
    printf("%c => ", *ptr);

    /* perform bitwise AND for every bit of the character */
    for(i = 7; i >= 0; --i) 
        if(*ptr & 1 << i) 
          str[7-i]='1';
        else
          str[7-i]='0';
        //(*ptr & 1 << i) ? putchar('1') : putchar('0');
    str[8]='\0';    
    printf("%s\n",str);
    memset(str,'/0',8);        
}
return 0;
}

Output:

H => 01001000
e => 01100101
l => 01101100
l => 01101100
o => 01101111
Abort trap

It would be nice if someone can throw some light. Even though i am getting the output, the trap is happening.

Courtesy: This is a modified program of a fellow stack fellow user @Athabaska.

Aasmund Eldhuset · Accepted Answer · 2011-04-15 18:02:09Z

3

str[8]='\0' causes a buffer overflow because str only has 8 elements, and you're trying to access the ninth one.

Edit: I complained about the '/0', but as @R.. informed me, it is valid - however, memset() will ultimately convert its parameter to unsigned char, so I don't think there is any point in using it, so I assume the author will want to change it to '\0'.

edited Apr 15, 2011 at 18:02

answered Apr 15, 2011 at 17:42

Aasmund Eldhuset

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3 Comments

R.. GitHub STOP HELPING ICE Over a year ago

'/0' is valid C. It's a multi-character constant which has implementation-defined value, usually defined as interpreting the bytes in that order as a native-endian number.

Aasmund Eldhuset Over a year ago

@R..: +1; I didn't know that. Don't see the point of passing a multi-character value to a function that will ultimately convert its parameter to 'unsigned char', though...

R.. GitHub STOP HELPING ICE Over a year ago

Well on typical implementations, it would fill with either '/' or '0' depending on the system's endianness. :-)

Andrey · Accepted Answer · 2011-04-15 17:41:21Z

3

str[8]='\0'; is invalid. If declaration is char str[8]; then valid indices are 0..7

answered Apr 15, 2011 at 17:41

Andrey

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Comments

Antti · Accepted Answer · 2011-04-15 17:42:21Z

2

str[] only has space for 8 characters, but you access the 9th: str[8] = '\0';

answered Apr 15, 2011 at 17:42

Antti

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