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Any got a solution for this programming challenge ?
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hi guys,
here is challenge i got from my friend,i just want to know there is a solution available for this task,
when i enter a number it should display it's equivalent binary number like for 2 = 0010 3=0011,
but the important restriction is "IT SHOULD NOT USE/INVOKE MULTIPLICATION OR DIVISION OF ANY FORM" inside the computer.
is there a solution possible ?
if any one has one please post here...
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Wrote a quick/dirty C program to illustrate a solution. Uses subtraction and logical AND operations. It converts hex number string to binary string.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void shell_prompt(void)
{
char buf[1024];
printf("Type exit to quit, valid string chars 0-9, A-F, a-f.\n");
while(1)
{
printf(">");
gets(buf);
if (strcmp(buf, "exit") == 0) break;
else string_convert(buf);
}
}
int string_convert(char *num)
{
char *ret;
int i;
ret = (char *)malloc(sizeof(char)*strlen(num));
/*
ASCII input -- 0 - 9: 48 - 57; A - F: 65 - 70; a - f: 97 - 102
*/
for (i = 0; i < strlen(num); i++)
{
if ((int)num[i] > 47 && (int)num[i] < 58)
ret[i] = num[i] - 48;
else if ((int)num[i] > 64 && (int)num[i] < 71)
ret[i] = num[i] - 55;
else if ((int)num[i] > 96 && (int)num[i] < 103)
ret[i] = num[i] - 87;
else
{
printf("Invalid character string. Valid characters 0-9, a-f, A-F.\n");
return 1;
}
if (ret[i] & 0x8) printf("1"); else printf("0");
if (ret[i] & 0x4) printf("1"); else printf("0");
if (ret[i] & 0x2) printf("1"); else printf("0");
if (ret[i] & 0x1) printf("1"); else printf("0");
printf(" ");
}
printf("\n");
return 0;
}
int main(int argc, char **argv)
{
if (argc < 2)
{
printf("No arguments given, defaulting to prompt.\n");
shell_prompt();
}
else string_convert(argv[1]);
return 0;
}
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Looks like the above code has a buffer overflow vuln, may be wrong but it appears that way.Skype: mrpt3oTwitter: MrPteo
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...this was only proof of concept, not designed to be secure, there probably is a buffer overflow vuln in here, but I'll leave that open for you to find out...
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haha was only looking through it as iv been learning buffer overflows recently.Skype: mrpt3oTwitter: MrPteo
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"Smashing the stack for Fun and Profit" is a good place to start... ;)
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hi bro,thanks for the solution mate :)
