solved that to... sorry for my lack of experience...
now i want to concatenate two bites from a byte to another byte. i have to use a macro
smthing like this
a: 10101011
b: 11101001
i want to obtain 1011101001 (take the first two from the begining of a and put them in the front of b) can i do this using a macro ?
thank you for your help!
You could for example mask out the 6 unwanted bit from a by using an appropriate bitmask and then shift the result two bits to the left and use bitwise OR to "add" b to the result.
You could also shift a 6 bits to the right, then shift 8 bits to the left and use bitwise OR to "add" b to the result. This will maybe cause a warning by the compiler.
This can of course be done with a makro.
i've done smthing like this because i only needed the decimal number.Code:#define C(x) (x>>7&1)*256+(x>>6&1)*128
next problem... i have a 4 byte entry that is an unsigned int, the problem is that i used a char to read the binary file. how can i obtain the 4 byte int ?
http://www.daniweb.com/code/snippet150.htmlCode:#include <stdio.h> #include <limits.h> static const char Bit[] = "01"; char *bits_uchar(char *dest, unsigned char value) { char *start = dest; unsigned char bit; for ( bit = UCHAR_MAX / 2 + 1; bit > 0; bit >>= 1 ) { *dest++ = Bit[!!(value & bit)]; } *dest = 0; return start; } char *bits_ushort(char *dest, unsigned short value) { char *start = dest; unsigned short bit; for ( bit = USHRT_MAX / 2 + 1; bit > 0; bit >>= 1 ) { *dest++ = Bit[!!(value & bit)]; } *dest = 0; return start; } int main(void) { unsigned char a = 0xAB, b = 0xE9; unsigned short c = ((a & 0xC0) << 2) | b; char binary[sizeof c * CHAR_BIT + 1]; printf("a = 0x%02X = %s\n", a, bits_uchar(binary, a)); printf("b = 0x%02X = %s\n", b, bits_uchar(binary, b)); printf("c = 0x%04X = %s\n", c, bits_ushort(binary, c)); return 0; } /* my output a = 0xAB = 10101011 b = 0xE9 = 11101001 c = 0x02E9 = 0000001011101001 */
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
My endianness link was a little vague.
Code:#include <stdio.h> int main(void) { unsigned char b[] = {0x12,0x34,0x56,0x78}; unsigned long value = (b[0] << 24) | (b[1] << 16) | (b[2] << 8) | b[3]; printf("value = 0x%08lX\n", value); return 0; } /* my output value = 0x12345678 */Results may vary depending on the endianness of the target.Code:#include <stdio.h> int main(void) { unsigned char b[] = {0x12,0x34,0x56,0x78}; unsigned long value = (b[3] << 24) | (b[2] << 16) | (b[1] << 8) | b[0]; printf("value = 0x%08lX\n", value); return 0; } /* my output value = 0x78563412 */
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
how can i do this viceversa, i have the long and i want the char [4]
i've tryed smthing like this but it doesn't work well:
Code:int main() { unsigned char a[5]={00000}; unsigned long x,i; printf("x=");scanf("%ld",&x); a[0]=a[0]|(x>>8); x=x>>8; a[1]=a[1]|(x>>8); x=x>>8; a[2]=a[2]|(x>>8); x=x>>8; a[3]=a[3]|(x>>8); for(i=0;i<4;i++) printf("%x ",a[i]); printf("\n"); return 0; }
Perhaps like this.Code:#include <stdio.h> #include <limits.h> void foo(unsigned long value) { unsigned char byte[sizeof value]; size_t i; for ( i = 0; i < sizeof byte / sizeof *byte; ++i ) { byte[i] = value >> (CHAR_BIT * i); printf("byte[%lu] = 0x%02X\n", (long unsigned)i, byte[i]); } } int main(void) { foo(0x12345678UL); return 0; } /* my output byte[0] = 0x78 byte[1] = 0x56 byte[2] = 0x34 byte[3] = 0x12 */
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
