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When overloading functions, it is possible to produce a type of error with which you may not be familiar You can create a situation in which the compiler is unable to choose between two (or more) overloaded functions When this happens, the situation is said to be ambiguous Ambiguous statements are errors, and programs containing ambiguity will not compile By far the main cause of ambiguity involves C++ s automatic type conversions C++ automatically attempts to convert the arguments used to call a function into the type of arguments expected by the function For example, consider this fragment:

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Hi Team, I want to upload pdf file and show it in the browser as it is. I was able to read pdf file using pdfbox but cannot display the content ...

As the comment indicates, this is not an error because C++ automatically converts the character c into its double equivalent In C++, very few type conversions of this sort are actually disallowed Although automatic type conversions are convenient, they are also a prime cause of ambiguity For example, consider the following program:

#include <iostream> using namespace std; float myfunc(float i); double myfunc(double i); int main() { cout << myfunc(101) << " "; // unambiguous, calls myfunc(double) cout << myfunc(10); // ambiguous return 0;

} float myfunc(float i) { return i; } double myfunc(double i) { return -i; }

#include <iostream> using namespace std;





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The LEX is connected to other LEX of ces through an interof ce trunk line Long-distance toll calls are connected to toll center of ces that connect to intertoll trunks Formerly, the interof ce and intertoll cable systems were copper, but most have been replaced by optical bers Regardless of the method, the telephone system network topology is the same This interconnection topology is shown in Figure 5-17

char myfunc(unsigned char ch); char myfunc(char ch); int main() { cout << myfunc('c'); // this calls myfunc(char) cout << myfunc(88) << " "; // ambiguous return 0; } char myfunc(unsigned char ch) { return ch-1;

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Here, myfunc( ) is overloaded so that it can take arguments of either type float or type double In the unambiguous line, myfunc(double) is called because, unless explicitly specified as float, all floating-point constants in C++ are automatically of type double Hence, that call is unambiguous However, when myfunc( ) is called using the integer 10, ambiguity is introduced because the compiler has no way of knowing whether it should be converted to a float or to a double This causes an error message to be displayed, and the program will not compile As preceding example illustrates, it is not the overloading of myfunc( ) relative to double and float that causes the ambiguity Rather, it is the specific call to myfunc( ) using an indeterminate type of argument that causes the confusion Put differently, the error is not caused by the overloading of myfunc( ), but by the specific invocation Here is another example of ambiguity caused by C++ s automatic type conversions:

Figure 5412 All connections are designed, but may be altered to suit the fabricator s preferences with relative ease prior to printing final shop drawings Embedded plates and the connections to them are in the model All material, including reinforcing at mechanical openings, is accounted for (Image courtesy of Gregory P Luth & Assoc, Inc)

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} char myfunc(char ch) { return ch+1; }

1 4 7 * 2 5 8 0 3 6 9 #

In C++, unsigned char and char are not inherently ambiguous However, when myfunc( ) is called by using the integer 88, the compiler does not know which function to call That is, should 88 be converted into a char or an unsigned char Another way you can cause ambiguity is by using default arguments in overloaded functions To see how, examine this program:

#include <iostream> using namespace std; int myfunc(int i); int myfunc(int i, int j=1); C++ int main() { cout << myfunc(4, 5) << " "; // unambiguous cout << myfunc(10); // ambiguous return 0; } int myfunc(int i) { return i; } int myfunc(int i, int j) { return i*j; }

Here, in the first call to myfunc( ), two arguments are specified; therefore, no ambiguity is introduced and myfunc(int i, int j) is called However, when the second call to myfunc( ) is made, ambiguity occurs because the compiler does not know whether to call the version of myfunc( ) that takes one argument or to apply the default to the version that takes two arguments

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Unable to open PDF files with Adobe Reader v11.0, in Windows 8 ...
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Viewing PDF in Windows forms using C# - Stack Overflow
i think the easiest way is to use the Adobe PDF reader COM Component. right click ... Reading/Writing PDF Files in Visual C# Windows Forms.












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