What is an unsigned int in C?
613866
its unsigned integer, as mentioned above its mostly ranged from 0 to 65,535 depends on your machine/compiler.\n\nnormally a signed integer can also be negative numbers, so an integer is ranged from -32,768 to +32,767\n\nread more about C data types here\nhttp://www.phim.unibe.ch/comp_doc/c_manual/C/CONCEPT/data_types.html
551498
I'm no "C" expert, but in Delphi this is a Windows system call, and I'm sure it will be a similar implementation in C; just alter the syntax as necessary...
114514
I have the answer.\n\nYou should change the numbers to be "400.0" instead of "400". This is your primary problem.\n\nThe deeper reason that this is happening is that 400*400/400 is an INTEGER expression. It will be evaluated as an integer, and only assigned to a floating point result.\n\nI believe that your compiler is a 16 bit compiler, this is the only answer that makes sense.\n\nNow let's just see what the compiler is doing. Its first step is to evaluate 400*400. Now if you get out your calculator, you will get the answer 160,000. NOT SO FAST!!!! An integer in your compiler is only 16 bits long, and can only express numbers between -32768 and 32767. The answer that it comes up with is 28,928 (160,000 - 2*65,536). It will next divide 28,928 by 400 to give 72 (again it drops the fractional result because it is integer).\n\nIn both C and C++, the result of an integer expression is an integer. It does not matter if it gets assigned to a float or not. If you want float accuracy, then the numbers in the expression must be floats.
1008669
sizeof doesnt give the length of what you actually have, its gives the size of how big the variable has been declared to be . . .\n\nsizeof will return the number of bytes reserved for a variable or data type. \n\nThe following code shows sizeof returning the length of a data type. \n\n\n        /* How big is an int? expect an answer of 4. */\n\nmain()\n {\n   printf("%d &#92;n", sizeof(int));\n        }\n\n \n\n\nsizeof will also return the number of bytes reserved for a structure. \n\n\n        /* Will print 8 on most machines. */\n\n        main()\n{\n  struct \n  {\n    int a;\n    int b;\n  } TwoInts;\n\nprintf("%d &#92;n", sizeof(TwoInts));\n}\n\n\n \n\n\nFinally, sizeof will return the length of a variable. \n\n\nmain()\n{\n  char String[20];\n  \n  printf ("%d &#92;n", sizeof String);\n   printf ("%d &#92;n", sizeof (String));\n        }\n\n\n \n\n\nIn the example above I have printed the size of 'String' twice. This is to show that when dealing with variables, the brackets are optional. I recommend that you always place the brackets around the sizeof argument.
365175
It actually refers to a memory location whose address\nin hex is 800H or in decimal it is 2048D.
361052
64-bit memory chip\n\n=D
1147441
Thesre is no operator ** in Cor other C based languages. Though it can be used with pointer to a pointer to get orginal value.\nSuch as\nint *p;\n*p=q;\nint **s;\n*s=p;\n**p=**p+5;//this will increase q by 5\n\n\n\nNow there is -- operator, that is decrement operator in C and all other C based language, which works just like ++.\n\nthat is like a++ increases a by 1, a-- decreases by1.
95584
I rember doing something like this: you had to take the int you store the input in, and divide it by ten with the mod operator (%), and store it in a new int.  Since the mod operator divides the two numbers, and then returns the remainder, that will separate your last number.  For example:\n\nSay the input is 4321, and you have it stored in int input;\n\nint first_num = input % 10;\n\nthis takes your value in input, divides it by 10, then takes the remainder and stores it in first_num.  In this case, first_num should now equal 1.  \n\nAfter that, you have to shorten your input from 4321 to 432 by dividing (regular dividing this time) it by ten, and storing it in another new int.  (Since int can't have any decimals, dividing by ten just shortens it by the first number).  Then you do the whole thing over until you have all four numbers in separate ints, and then output them to separate lines.  \n\nHope that's clear enough--it's hard to explain this stuff so it's easy to understand.  Good Luck!
121196
I understand you want to convert A,B,C... to {0,1}, right? then. In C, all strings are character arrays, otherwise collection of characters. by converting each character u can do. as u can see each character both printable and other have a numerical value. it is in a table, ASCII table. Here the code\n----------\nvoid binary(char ch)\n{  if(ch>0) binary(ch/2);\n   printf("%d",ch%2);\n}\nvoid toBinary(char*text)\n{  for(i=0;text[i]!='&#92;0';i++)\n   {  binary(test[i]);\n      printf("&#92;n");\n   }\n}\n----------\n\ni hope these two functions helps you.....\ngood luck.
711526
well, an integer is a whole number (postive or negative) or zero.\nsuch as 37463, or -2, or 0.
356128
Its alt+3, and FYI Alt and a number is a way of printing orginal symbols from ASCII text, (first printer data kinda thing)
523720
If you're asking how to convert a number (stored in binary)\ninto its decimal string, it really boils down to successive\ndivisions by 10, using the remainder as each digit (working\nfrom least significant to most).\n\nAn ugly but function version might be:\n\nchar *\nto_decimal_string(int v)\n{\n  char buff[STRSIZE];\n\n  int neg = v<0;\n  if (neg)\n    v=-v;\n\n  char *p = &buff[sizeof(buff)-1];\n  *p-- = 0;\n  if (v == 0)\n    return "0";\n  while (v>0) {\n    *p-- = "0123456789"[v%10];\n    v /= 10;\n  }\n  if (neg)\n    *p-- = '-';\n  return p+1;\n}\n\nIf you're asking how to convert an ascii string representing\na binary (base-2) number into another string representing\na decimal (base-10) number, I'd suggest using library\nroutines.  Any standard conforming C or C++ implementation\nwill provide "strtol()" and "sprintf()" (as well as other\nequivalents).\n\nThe following is purposefully done in the C subset of C++\nso that either compiler will compile it:\n\n#include <stdlib.h>\n#include <stdio.h>\n\nint\nmain(int argc, char *argv[])\n{\n  long value = strtol(argv[1], 0, 2);\n  printf("binary='%s', decimal='%d'&#92;n",\n    argv[1], value);\n  return 0;\n}\n\nThen to run it:\n\nThen to run it:\n\n$ CC convert.C\n$ a.out 10011101\nbinary='10011101', decimal='157'
881644
Something like this:\n\nint **myArray;\nint s1,s2,i;\n\nprintf("Input row size: ");\nscanf("%d",&s1);\nprintf("Input column size: ");\nscanf("%d",&s2);\n\nmyArray[0] = malloc(s1*s2*sizeof(int));\nif (myArray[0] != NULL ) {\n  for (i = 0;i<s1,i++) {\n         myArray[i] = (i * s2) + myArray[0];\n  }\n}\n\nmyArray can then be used as normal\n\nmyArray[2][4] = 20;
1033178
64 bits set in code unfortunately
662225
zero is an integer.\nAn integer is any whole number (positive or negative), including zero.
49715
use bitwise operators. something like:\n\nnum_bits = 0;\nfor(i = 0; i < 32; i++) {\nif number & (1<<i) num_bits++;\n}\n\nnow num_bits holds the number of bits that were set in number. The "<<" operator does bit shifting. So "1<<i" equals a number that's all 0's except for a 1 at position i. If you bitwise and that with something, that tells you if that bit is set in the number. So we do this for each of the bits and add up the results.\n\nThis is assuming 32-bit integers. Adjust accordingly for 64-bit integers.
117995
There are five data types for C: void, integer, float, double, and char.\n\nvoid: associated with no data type \nint: integer \nfloat :floating-point number \ndouble :double precision floating-point number \nchar: character \n\n\n\nC++ defines two more: bool and wchar_t.\n\nbool: Boolean value, true or false \nwchar_t: wide character