C++ :: Exceeded 32bit Register Value Exceeds 4 Bytes
May 28, 2014
My question, What happens when with the 32 bit register value exceeds 4 bytes while programming in c++? Im interested in encryption.
Im programming and doing some calculations. The hex value "0xFA062F2F" multiplied by "6D" sends it over "0x6A74A21703", which exceeds 32bit.
My problem is while programming in C++, I do some calculations such as above, and I get the hex value "0xC0000001" which should be "0x6A74A21703". I made sure that I use long long as my interger type.
How can I overcome this problem with programming.
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Mar 15, 2015
how to accommodate double-size:8 bytes in 4 bytes pointer in 32bit system.
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Mar 4, 2015
#include<iostream.h>
#include<conio.h>
void main() {
clrscr();
char name[5];
cout<<"Enter your name";
cin>>name;
cout<<"Your name is"<<name;
getch();
}
In the above program the size of the array of the variable name is 5. which means the variable cant store more than 5 characters.
which also means
If I give the string "LINISH"
It should only print LINIS
But while the program is running, Even if I type a 10 characters string, It is getting printed Completely..why?
#include<iostream.h>
#include<conio.h>
void main(){
clrscr();
char name[5];
[Code] .....
In this program when I input a string for the variable name,It is getting printed completely, irrespective how many characters are there in the string.But If the string (which is input to the second variable that is game )holds more than 5 characters. the input of the first variable(name) is getting disturbed..why?
look at the below cited output to be more clear about my doubts.
OUTPUT NO:1
Enter your name:LINISHFRANCIS (Note that the input holds more than five chars)
Enter your game:GOLF(input is less than five chars)
LINISHFRANCIS loves GOLF(Two inputs are getting printed comopletely)
OUTPUT NO:
Enter your name:LINISHFRANCIS (Note that the input holds more than five chars)
Enter your game:FOOTBALL(input is more than five chars)
ALL loves FOOTBALL [Note that "ALL" is the last three letters of FOOTBALL
I am using TurboC++ for windows 7
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Mar 24, 2013
Code:
int i12 = 1001;
cout << i12 << " " << &i12 << endl;
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It's 6 bytes, but I'd expect the address to be 8 bytes on my 64 bit machine. I am using Ubuntu 12.04, GNU compiler.
So, why the address is 6 bytes and not 8 bytes?
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Jan 31, 2013
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Code:
#include <stdio.h>
#include <stdlib.h>
int main()
{
int * vettore, n;
unsigned int len=1;
}
[code]....
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How to read pixels from an x,y coordinate using the VGA register values?
byte readVRAMdirect(VGA_Type *VGA, uint_32 start, uint_32 offset) //Used in 256 color mode!
{
if (!VGA->VRAM_size) return 0; //No size!
return VGA->VRAM[SAFEMOD((start*4)+offset,VGA->VRAM_size)]; //The full protected offset!
[code]....
getVRAMScanlineStart(VGA,y) gives the offset register multiplied by 2 (shl 1), multiplied with the current memory address size (byte, word or dword) multiplied by the scanline. 256-color mode is already working. GETBIT gives a bit #(0-7) of a specified byte.
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ZIP with screen captures: URL....
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Dec 7, 2014
I'm trying to make a cash register program, but first I need to reinitialize ones, fives, tens, and twenties to their correct dollar amount. However, whenever I try to pass the user inputted values to the function that calculates the correct dollar amount, it just prints the same number.
Here's my code:
#include <iostream>
#include <string>
using namespace std;
//dollarAMT - recalculates the actual dollar amount
void dollarAMT(int ones, int fives, int tens, int twenties){
[Code] ....
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Dec 7, 2014
I'm trying to make a cash register program, but first I need to reinitialize ones, fives, tens, and twenties to their correct dollar amount. However, whenever I try to pass the user inputted values to the function that calculates the correct dollar amount, it just prints the same number.
Here's my code:
#include <iostream>
#include <string>
using namespace std;
//dollarAMT - recalculates the actual dollar amount
void dollarAMT(int ones, int fives, int tens, int twenties){
[Code] ....
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Oct 23, 2013
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Nov 2, 2012
How many bytes in a string ....
How many bit in a string ...
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Jan 26, 2014
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#include <inttypes.h>
typedef uint64_t byte_int_t;
int main(void){
byte_int_t t;
printf("%" PRIu64 "
", t);
}
./runprogram
13217906525252912201
However, when I use a calculator, I get a different value: 2^64= 1.8446744e+19
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#include <stdio.h>
#include <inttypes.h>
typedef uint64_t byte_int_t;
int main(void) {
byte_int_t t;
printf("%u
", sizeof(t));
return 0;
}
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Code:
#include <iostream>
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#include <stdlib.h>
#include <windows.h>
using namespace std;
int main() {
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[Code] .....
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This happens on all the image files I have tested, and furthermore, on several other non-image files. They all start printing out ' ' after a while. However, all these files are fine and not corrupted, e.g. the image files display correctly.
Code:
#include <fstream>
#include <iostream>
#include <sstring>
#include <string>
#include <stdio.h>
using namespace std;
string CharToBin(char ch) {
bool bits[8];
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Sep 4, 2012
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The problem is, when I convert "c" to "c_int", it adds on a load of 1's to the left of the "c" data. This then means that the OR comparison changes all those bits in my integar to 1.
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Code:
int StreamFileToInt(FILE *fp, int num_bytes) {
char c;
int c_int;
int x = 0x0000;
for (int i = 0; i < num_bytes; i++) {
[Code] ....
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lun, 08 julio 2013 04:28:17 UTC <---
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Feb 22, 2013
while bytesremaining >=8 do something;
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Feb 22, 2013
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May 13, 2013
I made this program to convert bits to bytes, because I'm so sick of seeing ISP's advertise speeds in megabits, which I consider an intentional attempt to decieve :P And I think I've finally understood how the return value of scanf works since the last time I posted here, so my program can check to see if an integer was entered before processing the input, but I'm stuck on how to make the whole program start over if an integer is not entered. I have a hunch it would involve a loop, but I can't figure out how to make the program start over at "How many mb do you need converted?" if an integer is not entered into scanf..Here is the code I have so far:
Code:
#include <stdio.h>
int main () {
int b, mb, kb, Byte, kB, mB, gB;
char term;
}
[code]....
and my program makes the assumption for now at least, that mb will be inputted because that's the unit of measurement that i usually see advertised, and i didn't bother making an if statement to print a conversion in terms of gigabytes because i've never heard of a connection that fast :P
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May 26, 2013
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Jan 6, 2015
how to combine two bytes (each byte is 8 bits wide) in order to get the original value. I am displaying values(from 0 to 500)sent from my microcontroller on the GUI.I can successfully display values from 0 to 255 as this requires just sending a byte. However sending values from 256 to 500 requires sending two bytes. The problem I am having is that I was unable to re-combine the received two bytes in order to get the original value. Below are my lines of code:
int main( void ) // this is the main function inside the microcontroller {
sei();
USI_TWI_Master_Initialise();
Spi_Master_Init();
while(1) // I am using this loop to send 500(111110100) {
Transmitt_Receive(244);//this function sends LOW BYTE of 500(11110100)
_delay_ms(1000);
[code]....
When I combined the two bytes using the above arrangements, instead of getting 500 my GUI displayed 62708.I got the same result when I used BitConverter.
value3 = BitConverter.ToInt32(val,0);
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