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Getting widthhight of an EPS



Getting widthhight of an EPS

#$imgdata[1] contains width /[2] contains height 


$fp=fopen ($img, "r");
$buffer = fgets($fp, 4096);
if (preg_match("/ImageData:[^\"]*\"/",$buffer ,$imgdataln)) {			  
  $imgdata=split(" ", $imgdataln[0]); 
}
fclose ($fp)

Number Decimal System to Binary System
C Language program code to using recursion finds a binary equivalent of a decimal number entered by the user. The user has to enter a decimal which has a base 10 and the program

C++ Detecting a Keypress and ASCII Code
Program which detects a key pressed and its 'ASCII' value with out pressing enter key and using cin>> on press Esc key program should exit. When press a key its ASCII value should

Program Sortings an Array of 10 Elements
"Heapsort" can be thought of as an improved "selection sort": like that algorithm, it divides its input into a sorted and an unsorted region, and it iteratively shrinks the unsorted region

C++ Code Solves Tower of Hanoi Problem
In The C++ program, displays the solution to the Tower of Hanoi problem using the base-2 representation of the move number with the following rules:- There is one binary digit (bit)

Simple Singly Linked List Programs in C++
In C++, Linked list is a linear collection of data elements, called nodes, each pointing to the next node by means of a pointer. So this is a data structure consisting of a group of nodes

Program to implement a boolean function
Color at stars position. Start in upper corner. Gives a message to user. For checking that a particular no. Decides whether output is 1 or 0. Normal mux implementation. Decides...

Count Occurrence Positive Zero Negative
We have to enter some set of numbers. Now to find occurrence of positive, negative, zero from the given set of numbers, just check all the numbers using for loop whether number

Code find maximum using ladder if.else.if
Instead of using Nested If Else. Lets combine 2 or more conditions together using Logical Operators. A number j1 among 3 numbers j1, j2 and j3 is said maximum if j1 > j2 and j1 > j3.