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I am trying to research various integer overflow scenarios in C and I was wondering does the C language provide any defenses against numeric errors and are there any additional classes or libraries in the C language that can help with that? Also, can anyone give me an example of code that results in an integer overflow in C?
No, there are no defenses.
This overflows:
#include <limits.h>
#include <stdio.h>
const int a = INT_MAX - 2;
const int b = INT_MAX - 2;
printf("%d + %d = %d\n", a, b, a + b);
When I tested it it printed -6, but anything could happen I guess.
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I need to understand more about this if condition the two sides of comparison and how it is compared:
int main()
{
unsigned short i;
if (i == '9' * 256 + '5')
{
/* Do stuff */
}
}
How are these compared?
Formally the behaviour of your code is undefined as you are reading the uninitialised variable i.
'9', 256, and '5' are all int types in C. So the right hand side is evaluated in int arithmetic, with the potential for overflow (it will not overflow with ASCII encoding).
i will be converted to an int type prior to the comparison.
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Can I make a float number in a C program always round up
You can use the ceil() function. For example:
#include <stdio.h>
#include <math.h>
int main () {
float val1 = 1.6;
printf ("Round up to %.1lf\n", ceil(val1));
return(0);
}
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I just wonder how to find mean of two numbers without using division.
do not use these conditions :
int mean = (a + b) >> 1;
four fundamental arithmetic operations
I think this may be helpful -->
int a,b,i,j;
if (a>b)
{
int temp = a;
a = b;
b = temp;
}
for(i=a,j=b;i<j;i++,j--)
continue;
if(i==j)printf("%d\n", i);
else printf("%lf\n", (double)(i)-0.5);
Add them then multiply by 0.5 , no division involved.
If they're both integers, you can use a right shift:
int median = (a + b) >> 1;
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Below is a C program which wants to multiply an integer by 5 using bitwise operation. But when i run this program it gives unexpected output. I know there is something i messing up which i could not see. Any help regarding this guys and girls ?
#include <stdio.h>
#define PrintInt(expr) printf("%s : %d\n",#expr,(expr))
int FiveTimes(int a)
{
int t;
t = a<<2 + a;
return t;
}
int main()
{
int a = 1, b = 2,c = 3;
PrintInt(FiveTimes(a));
PrintInt(FiveTimes(b));
PrintInt(FiveTimes(c));
return 0;
}
This is a question of "operator precedence": "<<" has a lower priority than "+" - so your code actually calculates a << (2 + a), while it should be (a << 2) + a. The latter is the fix.
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Need to write a C program that adds two numbers each of 100+ digits..
I don't want the method of using arrays to do this.
Please suggest me how to store this numbers(atleast of 512 bit sized) and do the arithmetic operations?
You could use an arbitrary precision arithmetic library, such as GMP for that.
A quick C example:
#include <gmp.h>
mpz_t a, b;
const char *huge_decimal_num1 = "46819294521564960351683095841209562359068";
const char *huge_decimal_num2 = "6904120584864540916814056801234572451249681";
mpz_init_set_str (a, huge_decimal_num1, 10);
mpz_init_set_str (b, huge_decimal_num2, 10);
mpz_add (a, a, b); // a = a + b
printf("%s + %s = %s\n",
huge_decimal_num1, huge_decimal_num2, mpz_get_str (NULL, 10, a));