How to convert Char array to long in obj c
unsigned char *composite[4];
composite[0]=spIndex;
composite[1]= minor;
composite[2]=shortss[0];
composite[3]=shortss[1];
i need to convert this to Long int..Anyone please help
If you are looking at converting what is essentially already a binary number then a simple type cast would suffice but you would need to reverse the indexes to get the same result as you would in Java: long value = *((long*)composite);
You might also consider this if you have many such scenarios:
union {
unsigned char asChars[4];
long asLong;
} value;
value.asChar[3] = 1;
value.asChar[2] = 9;
value.asChar[1] = 0;
value.asChar[0] = 10;
// Outputs 17367050
NSLog(#"Value as long %ld",value.asLong);
Related
I want to cast logic packed array into longint unsigned in systemverilog and then I can export it using DPI-C to C++ unsigned long. The simulator I am using is Verilator. Check the example below.
logic[31:0] v1;
logic[63:0] v2;
int a = signed'(v1); //cast to signed int
int b = int'(v1); //cast to signed int
int unsigned c = unsigned'(v1); //cast to unsigned int
longint d = longint'(v2); //cast to signed long
//longint unsigned e = longint unsigned'(v2); //This doesn't work. I need to cast to unsigned long.
You need to create a SystemVerilog type without a space in it using typedef. Here's an example:
// ..
typedef longint unsigned uint64_t;
uint64_t e = uint64_t'(v2);
There's no need for any kind of cast unless sign extension is required. There is already implicit casting between 4-state and 2-state types.
You can just write:
longint d = v2;
#include <stdio.h>
int main(void){
char c[8];
*c = "hello";
printf("%s\n",*c);
return 0;
}
I am learning pointers recently. above code gives me an error - assignment makes integer from pointer without a cast [enabled by default].
I read few post on SO about this error but was not able to fix my code.
i declared c as any array of 8 char, c has address of first element. so if i do *c = "hello", it will store one char in one byte and use as many consequent bytes as needed for other characters in "hello".
Please someone help me identify the issue and help me fix it.
mark
i declared c as any array of 8 char, c has address of first element. - Yes
so if i do *c = "hello", it will store one char in one byte and use as many consequent bytes as needed for other characters in "hello". - No. Value of "hello" (pointer pointing to some static string "hello") will be assigned to *c(1byte). Value of "hello" is a pointer to string, not a string itself.
You need to use strcpy to copy an array of characters to another array of characters.
const char* hellostring = "hello";
char c[8];
*c = hellostring; //Cannot assign pointer to char
c[0] = hellostring; // Same as above
strcpy(c, hellostring); // OK
#include <stdio.h>
int main(void){
char c[8];//creating an array of char
/*
*c stores the address of index 0 i.e. c[0].
Now, the next statement (*c = "hello";)
is trying to assign a string to a char.
actually if you'll read *c as "value at c"(with index 0),
it will be more clearer to you.
to store "hello" to c, simply declare the char c[8] to char *c[8];
i.e. you have to make array of pointers
*/
*c = "hello";
printf("%s\n",*c);
return 0;
}
hope it'll help..:)
I have a slice of bytes (which I know that are an integer saved as little endian) and I want to convert them to an integer.
When I had a static-sized array it was no problem, but now I have a slice (ubyte[]).
Is it possible to still convert it to an integer, e.g. in this fashion?
ubyte[] bytes = ...;
uint native = littleEndianSliceToNative!uint(bytes);
Taking further what Adam has written, you can write a simple function like
T sliceToNative(T)(ubyte[] slice) if(isNumeric!T) {
const uint s = T.sizeof,
l = min(cast(uint)s, slice.length);
ubyte[s] padded;
padded[0 .. l] = slice[0 .. l];
return littleEndianToNative!T(padded);
}
You could even make the littleEndianToNative a generic type too so you mirror all the operations on arrays for slices.
Just slice the slice explicitly to the appropriate size:
import std.bitmanip;
void main() {
ushort i = 12345;
ubyte[2] swappedI = nativeToLittleEndian(i);
ubyte[] slice = swappedI;
alias Target = ushort; // make this a template param for a generic function
assert(i == littleEndianToNative!Target(slice[0..Target.sizeof])); // the [0..Target.sizeof]
}
That should work for any size needed.
I have a char array representing a binary number for example
bit <1x8 char> '00110001'
I want to replace the last char with a logical value. The following error is triggered: Conversion to char from logical is not possible.
This is my code:
bit(end:end) = hiddenImg(i,j);
I checked that hiddenImg(i,j) is in fact a logical value.
This may not be optimal but should do what you want (convert the logical to a char):
>> bit = '10010100'
bit =
10010100
>> bit(end)=num2str(true)
bit =
10010101
not sure why this is so difficult for me. I have an array and I store its count like so:
int index = myarray.count;
then I have a for loop like so and I simply want to add index and my i counter below:
for (int i=0; i < anotherarray.count ; i++) {
int newIndex = index + i;
}
this gives me the error:
assignment makes integer from pointer without a cast
and a crazy value
Note that index is a char * function, defined in string.h