This code show this error Invalid operands to binary expression ('int*' and 'int*')
int *a = 5;
int *b = 3;
int *c = a*b;
How I can multiply them, and Why is this a error?
you are initializing int* with an int value. To multiply values and write them to another pointer, you should call the value of a stored in an address with (*a)
here is the example code for what you want to do:
int *a = malloc(sizeof(int));
int *b = malloc(sizeof(int));
int *c = malloc(sizeof(int));
*a = 2;
*b = 3;
*c = (*a)*(*b);
printf("%d %d %d", *a,*b,*c);
free(a);
free(b);
free(c);
it prints 2 3 6 as expected.
this is the C (primitive type) Integer type so you cannot declare with pointer.
Please find the below code :
int a = 5;
int b = 3;
int c = (a*b);
Thanks
Please remove asterisk symbol because int is a primitive data type not an object. You can put asterisk symbol with objects not primitive data types.
Asterisk symbol is pointer symbol from the C language.
int a = 5;
int b = 3;
int c = a*b;
Related
I am facing problem with the objective c code to convert decimal to binary. When I enter small values it shows me the output.
For e.g. 12 -> 1010
But when I enters large numbers, it shows me the output as "10..." (includes dots in the output)
Please help me.
My program is as follows:
NSUInteger x = [newDec integerValue];
//int y[30];
int i=0;
int m =1;
while (x != 0) {
int mod = x % 2;
x /= 2;
i = i + mod * m;
m = m * 10;
string = [NSString stringWithFormat:#"%d", i];
}
There are two problems with your code.
1) Your label size is perhaps not able to accommodate your string. So check the length of it.
2) Your code will not support the conversion if value of x is large. The reason is that int has limited capacity. Check this question regarding memory size of in-built variable. So, consider making your string mutable and add 0s or 1s in it. I am attaching my snippet of code.
NSMutableString *string = [[NSMutableString alloc] init];
while (x != 0) {
int mod = x % 2;
x /= 2;
[string insertString:[NSString stringWithFormat:#"%d", mod] atIndex:0];
}
NSLog(#"String = %#", string);
I've prototyped an algorithm for my iOS game in Python, and I need to rewrite in in ObjC. Basically, I have a board of 16 numbers, and I want to loop through every number three times and the four functions I'm using (add, subtract, multiply, exponentiate). 1+2+3, 2*3-4, 3^4-5, 9-4^3, etc., but without order of operations (first operation is always done first).
What I would like is an overview of how this might be implemented in Objective-C. Specifically, what is the equivalent of an array of functions in Objective-C? Is there an easy way to implement it with selectors? What's the best structure to use for loops with numbers? Array of NSIntegers, array of ints, NSArray/NSMutableArray of NSNumbers?
import random as rand
min = 0
max = 9
max_target = 20
maximum_to_calculate = 100
def multiply(x, y):
return x * y
def exponate(x, y):
return x ** y
def add(x, y):
return x + y
def subtract(x, y):
return x - y
function_array = [multiply, exponate, add, subtract]
board = [rand.randint(min, max) for i in xrange(0, 16)]
dict_of_frequencies = {}
for a in board:
for b in board:
for first_fun in function_array:
first_result = first_fun(a, b)
for c in board:
for second_fun in function_array:
final_result = second_fun(first_result, c)
if final_result not in dict_of_frequencies:
dict_of_frequencies[final_result] = 0
dict_of_frequencies[final_result] += 1
The most convenient way in Objective-C to construct an array of functions would be to use Blocks:
typedef NSInteger (^ArithmeticBlock)(NSInteger, NSInteger);
ArithmeticBlock add = ^NSInteger (NSInteger x, NSInteger y){
return x + y;
};
ArithmeticBlock sub = ^NSInteger (NSInteger x, NSInteger y){
return x - y;
};
NSArray * operations = #[add, sub];
Since there's no great way to perform arithmetic on NSNumbers, it would probably be best to create and store the board's values as primitives, such as NSIntegers, in a plain C array. You can box them up later easily enough, if necessary -- #(boardValue) gives you an NSNumber.
If you want to do it with straight C function pointers, something like this will do it:
#include <stdio.h>
#include <math.h>
long add(int a, int b) {
return a + b;
}
long subtract(int a, int b) {
return a - b;
}
long multiply(int a, int b) {
return a * b;
}
long exponate(int a, int b) {
return pow(a, b);
}
int main(void) {
long (*mfunc[4])(int, int) = {add, subtract, multiply, exponate};
char ops[4] = {'+', '-', '*', '^'};
for ( int i = 0; i < 4; ++i ) {
printf("5 %c 9 = %ld\n", ops[i], mfunc[i](5, 9));
}
return 0;
}
and gives the output:
paul#MacBook:~/Documents/src$ ./rndfnc
5 + 9 = 14
5 - 9 = -4
5 * 9 = 45
5 ^ 9 = 1953125
paul#MacBook:~/Documents/src$
Function pointer syntax can be slightly convoluted. long (*mfunc[4])(int, int) basically translates to defining a four-element array, called mfunc, of pointers to functions returning long and taking two arguments of type int.
Maddy is right. Anyway, I'll give it a try just for the fun of it.
This has never seen a compiler. So please forgive me all the typos and minor syntax errors in advance.
#include <stdlib.h>
...
const int MIN = 0;
const int MAX = 9;
const int MAX_TARGET = 20;
const int MAX_TO_CALCULATE = 100;
...
- (int) multiply:(int)x with:(int)y { return x * y; }
- (int) exponate:(int)x with:(int)y { return x ^ y; }
- (int) add:(int)x to:(int)y { return x + y; }
- (int) substract:(int)x by:(int)y { return x - y; }
// some method should start here, probably with
-(void) someMethod {
NSArray *functionArray = [NSArray arrayWithObjects: #selector(multiply::), #selector(exponate::), #selector(add::), #substract(multiply::), nil]; // there are other ways of generating an array of objects
NSMutableArray *board = [NSMutableArray arrayWithCapacity:16]; //Again, there are other ways available.
for (int i = 0; i < 16; i++) {
[board addObject:#(arc4random() % (MAX-MIN) + MIN)];
}
NSMutableDictionary dictOfFrequencies = [[NSMutableDictionary alloc] init];
for (NSNumber a in board)
for (NSNumber b in board)
for (SEL firstFun in functionArray) {
NSNumber firstResult = #([self performSelector:firstFun withObject:a withObject:b]);
NSNumber countedResults = [dictOfFrequencies objectForKey:firstResult];
if (countedResults) {
[dictOfFrequencies removeObjectForKey:firstResult];
countedResults = #(1 + [countedResults intValue]);
} else {
countedResults = #1; // BTW, using the # followed by a numeric expression creates an NSNumber object with the value 1.
}
[dictOfFrequencies setObject:countedResults forKey:firstResult];
}
}
Well, let me add some comments before others do. :-)
There is no need for objective c. You python code is iterative therefore you can implement it in plain C. Plain C is available where ever Objective C is.
If you really want to go for Objective-C here then you should forget your python code and implement the same logic (aiming for the same result) in Objective-C in an OOP style. My code really tries to translate your code as close as possible. Therefore my code is far far away from neither beeing good style nor maintainable nor proper OOP. Just keep that in mind before you think, ObjC was complicated compared to python :-)
I want to convert some bytes to an int.
This is my code so far:
unsigned char *bytePtr = (unsigned char *)[aNSDataFrame];
I want to take 4 bytes from this unsigned char:
myFrame[10], myFrame[11], myFrame[12] and myFrame[13] and convert them to an integer.
int val = *(const int*)&myFrame[10];
you can do,
int a;
a=myframe[10];
a=a<<8;
a=a|myframe[11];
a=a<<8;
a=a|myframe[12];
a=a<<8;
a=a|myframe[13];
this will create integer containing those bytes
int bytesToInt(unsigned char* b, unsigned length)
{
int val = 0;
int j = 0;
for (int i = length-1; i >= 0; --i)
{
val += (b[i] & 0xFF) << (8*j);
++j;
}
return val;
}
Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
This question does not appear to be about programming within the scope defined in the help center.
Closed 8 years ago.
Improve this question
I am working on a app which sends data to server with user location info. Server accept this data based on checksum calculation, which is written in java.
Here is the code written in Java:
private static final String CHECKSUM_CONS = "1217278743473774374";
private static String createChecksum(double lat, double lon) {
int latLon = (int) ((lat + lon) * 1E6);
String checkSumStr = CHECKSUM_CONS + latLon;
byte buffer[] = checkSumStr.getBytes();
ByteArrayInputStream bais = new ByteArrayInputStream(buffer);
CheckedInputStream cis = new CheckedInputStream(bais, new Adler32());
byte readBuffer[] = new byte[50];
long value = 0;
try {
while (cis.read(readBuffer) >= 0) {
value = cis.getChecksum().getValue();
}
} catch (Exception e) {
LOGGER.log(Level.SEVERE, e.getMessage(), e);
}
return String.valueOf(value);
}
I tried looking for help to find out how to write objective c equivalent of this. Above function uses adler32 and I don't have any clue about that. Please help.
Thanks for your time.
The answers shown here by #achievelimitless and #user3275097 are incorrect.
First off, signed integers should not be used. The modulo operator on negative numbers is defined differently in different languages, and should be avoided when possible. Simply use unsigned integers instead.
Second, the loops will quickly overflow the 16-bit accumulators, which will give the wrong answer. The modulo operations can be deferred, but they must be done before overflow. You can calculate how many loops you can do safely by assuming that all of the input bytes are 255.
Third, because of the second point, you should not use 16-bit types. You should use at least 32-bit types to avoid having to do the modulo operation very often. You still need to limit the number of loops, but the number gets much bigger. For 32-bit unsigned types, the maximum number of loops is 5552. So the basic code looks like:
#define MOD 65521
#define MAX 5552
unsigned long adler32(unsigned char *buf, size_t len)
{
unsigned long a = 1, b = 0;
size_t n;
while (len) {
n = len > MAX ? MAX : len;
len -= n;
do {
a += *buf++;
b += a;
} while (--n);
a %= MOD;
b %= MOD;
}
return a | (b << 16);
}
As noted by #Sulthan, you should simply use the adler32() function provided in zlib, which is already there on Mac OS X and iOS.
On basis of definition of adler32 checksum as mentioned in wikipedia,
Objective C implementation would be like this:
static NSNumber * adlerChecksumof(NSString *str)
{
NSMutableData *data= [[NSMutableData alloc]init];
unsigned char whole_byte;
char byte_chars[3] = {'\0','\0','\0'};
for (int i = 0; i < ([str length] / 2); i++)
{
byte_chars[0] = [str characterAtIndex:i*2];
byte_chars[1] = [str characterAtIndex:i*2+1];
whole_byte = strtol(byte_chars, NULL, 16);
[data appendBytes:&whole_byte length:1];
}
int16_t a=1;
int16_t b=0;
Byte * dataBytes= (Byte *)[data bytes];
for (int i=0; i<[data length]; i++)
{
a+= dataBytes[i];
b+=a;
}
a%= 65521;
b%= 65521;
int32_t adlerChecksum= b*65536+a;
return #(adlerChecksum);
}
Here str would be your string as mentioned in your question..
So when you want to calculate checksum of some string just do this:
NSNumber * calculatedChkSm= adlerChecksumof(#"1217278743473774374");
Please Let me know if more info needed
The following produces the below error:
int calc_ranks(ranks)
{
double multiplier = .5;
return multiplier * ranks;
}
The return type double is not a int, as defined by the method calc_ranks. How do I round/cast to an int?
Round it using the round() method:
int calc_ranks(ranks) {
double multiplier = .5;
return (multiplier * ranks).round();
}
You can use any of the following.
double d = 20.5;
int i = d.toInt(); // i = 20
int i = d.round(); // i = 21
int i = d.ceil(); // i = 21
int i = d.floor(); // i = 20
You can simply use toInt() to convert a num to an int.
int calc_ranks(ranks)
{
double multiplier = .5;
return (multiplier * ranks).toInt();
}
Note that to do exactly the same thing you can use the Truncating division operator :
int calc_ranks(ranks) => ranks ~/ 2;
I see a lot of answers, but with less description. Hope my answer will add some value.
Lets initalize the variable, and see how it will change with different methods.
double x = 8.5;
toInt()
It truncates the decimal value.
int a = x.toInt();
print(a); // 8
truncate()
It also truncates the decimal value.
int b = x.truncate();
print(b); // 8
round()
It returns the closest integer. It uses half up rounding mode.
int c = x.round();
print(c); // 9
ceil()
It returns the closest integer greater than the value.
int c = x.ceil();
print(c); // 9
floor()
It returns the closest integer smaller than the value.
int c = x.floor();
print(c); // 8
I looked at the answers above and added some more answers to make it a little easier to understand.
double value = 10.5;
Using toInt()
void main(){
double value = 10.5;
var y = value.toInt();
print(y);
print(y.runtimeType);
}
Using round()
The round() method returns the closest integer to the double.
void main(){
double value = 9.6;
var b = value.round();
print(b);
print(b.runtimeType);
}
Using ceil()
The ceil() method returns the smallest integer that is equal or greater than the given double.
void main(){
double value = 9.5;
var d = value.ceil();
print(d);
print(d.runtimeType);
}
Using floor()
The floor() method returns the greatest integer not greater than the given double.
void main(){
double value = 10.9;
var j = value.floor();
print(j);
print(j.runtimeType);
}
Conclusion
We’ve gone through 4 different techniques to convert a double to an integer in Dart and Flutter. You can choose from the method that fits your use case to solve your problem. Flutter is awesome and provides a lot of amazing features.
To convert double to int just this:
division
double01 ~/ double02
PS: The operator x ~/ y is more efficient than (x / y).toInt().
Multiplication, addition and subtraction
(double01 * double02).toInt
(double01 + double02).toInt
(double01 - double02).toInt
Its easy,
(20.8).round()
For String,
double.tryParse(20.8).round()
from string to int ->
if you string in int format like '10'
then use ---->
int.parse(value)
but if string in double format like '10.6'
then use like this ---->
double.parse(value).toInt()
convert double to int
doubleValue.toInt()
Try this!
int calc_ranks(ranks)
{
double multiplier = .5;
return (multiplier * ranks).truncate();
}
class CurrencyUtils{
static int doubletoint(double doublee) {
double multiplier = .5;
return (multiplier * doublee).round();
}
}
----------------------
CustomText( CurrencyUtils.doubletoint(
double.parse(projPageListss[0].budget.toString())
).toString(),
fontSize: 20,
color: Colors.white,
font: Font.QuicksandSemiBold,
),
There's another alternative, you can first cast the double to 'num' datatype and then convert to int using toInt().
double multiplier = .5;
return ((multiplier * ranks) as num).toInt();
The num type is an inherited data type of the int and double types.
You can cast both int and double to num, then cast it again to whatever you want
(double -> use toDouble(), int -> use toInt())