a  programe  in c++ that    check  weather  a  number  is  integer  or  not?
345718
int main (void)\n{\nint a,rem\ncout<<"enter the number";\n<<cin a;\nrem=a/a;\nif(rem>0)\ncout<<"its integer";\nelse\ncout<<"not a integer";\nreturn 0;\n}
101861
The Integer class includes a method named parseInt() that serves just this purpose. An example of a simple program that performs this conversion is shown below:\n\n\npublic class s2i {\n\n   public static void main (String[] args) {\n\n      // String s = "fred";    // do this if you want an exception\n\n      String s = "100";\n\n      try {\n         int i = Integer.parseInt(s.trim());\n         System.out.println("int i = " + i);\n      } catch (NumberFormatException nfe) {\n         System.out.println("NumberFormatException: " + nfe.getMessage());\n      }\n\n   }\n\n}\n\n\n\nOther notes\n\nInteger.toString(int i) is used to convert in the other direction, from int to String. \n  \nIf you're interested in converting a String to an Integer object, use the valueOf() method of the Integer class instead of the parseInt() method.
667791
Do you mean "not equal to"?\n<>\nThat's also still supported in some newer languages, although != is more common nowadays.
1120168
Equals another negative number
497029
No, because the limit as x->2- is undefined.  The limit exists for x->2+.
846187
if you are using a "while" or "do while" loop\n\nwhile ((x1 - x0) > 0.001)\n{\n//your code here\n}\n\nor you can do this\n\ndo\n{\n//your code here\n}\nwhile ((x1 - x0) > 0.001);
966120
try this stuff... pls understand the logic...\n\n#include<stdio.h>\n\nint main()\n{\nint no,rem;\nscanf("%d",&no);\nwhile (no)\n{\nrem = no % 10;\nprintf("%d",rem);\nno = no / 10;\n}\n}
751233
I'm not sure why you are avoiding the loop, but you could use recursion instead. If I understand your question correctly this function in c++ should give you your desired results.\n\nint sum(int nbr){\n    if(nbr > 0){\n           return(nbr%10 + sum(nbr/10));       \n    }\n    else{\n         return(0);\n    }\n}\n\nIf you called it with a number such as 256, it would return an answer of 13, or the sum of each digit in the integer.
734599
kbhit will return a integer to you, go to asciitable.com to find out what des each interger corresponds to what key.\n\ngood luck
121013
if it is in binary, 1 + 1 = 10
330547
#include<math.h>\nthen, pow(10,i)\nwill give the antilog of i.
676530
In c++\n\n#include <iostream.h>\n\nint main(int argc, char* argv[])\n{\n printf("Hello World!&#92;n");\n return 0;\n}
406488
If the number in front of x^2 is positive it opens up.\nIf it is negative it opens down.
654083
a declaration\n\nexample:\nint count; defines count as an integer
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.
604143
here is a program for 10 numbers in ascending order\n\n#include <iostream.h>\n#include <conio.h>\n\nvoid main()\n{\nclrscr();\nint array[10],t;\nfor(int x=0;x<10;x++)\n{\ncout << "Enter Integer No. " << x+1 << " : " << endl;\ncin>>array[x];\n}\nfor (x=0;x<10;x++)\n{\nfor(int y=0;y<9;y++)\n{\nif(array[y]>array[y+1])\n{\nt=array[y];\narray[y]=array[y+1];\narray[y+1]=t;\n}\n}\n}\ncout << "Array in ascending order is : ";\nfor (x=0;x<10;x++)\ncout << endl << array[x];\ngetch();\n}\n\nThis program takes in the ten integers array[x] to be stored in the single-dimensional array as a screen input from the user.\nIt then sorts out these ten integers into ascending order and prints them out using the 'cout' command.
444754
This sounds like school work, its a fairly simple program. you will probably want to use the cout and cin function which needs you to include iostream.h \n\nsimply use cout to print to screen, ask for input etc. then take the input using cin and store it to a int variable.\n\na good way to do it would be to ask the user what they wish to do, eg subtract, divide and take that operator first, a symbol or a number to show choice. then the next line asks for the numbers to be input.\n\neg:\n\n---------\n\nHello. what do you wish to do:\n1 - subtract\n2 - divide\n3 - etc etc\n\n>\n\n-------------\n\nthen use an if function or maybe a switch to check what their input was and do the artithmatic.\n\nso a very basic prgram may look like this:\nit may have errors i havant checked it.\n\n-----------------\n\n#include <iostream>\n\nint main() \n{\nint number;\n\n    cout << "Hello, input a number" << endl;\n    cin >>number;\n    return 0;\n}\n\n------\n\nEDIT; i re read the req, you might find inputting the selection to a char better, then check the char to see what symbol is stored in it, eg + / -. and maybe use a char array for the number input. simple take the char array, check for the space. split the numbers into 2 and caste then to an int. sounds complicated but it isnt to hard.\n\nsorry about spelling mistakes.