C – How to Display Hexadecimal Numbers

c++formathexintegerprintf

I have a list of numbers as below:

0, 16, 32, 48 …

I need to output those numbers in hexadecimal as:

0000,0010,0020,0030,0040 …

I have tried solution such as:

printf("%.4x", a); // Where a is an integer

But the result that I got was:

0000, 0001, 0002, 0003, 0004 …

I think I'm close there. How can I fix it?

I'm not so good at printf in C.

Best Answer

Try:

printf("%04x",a);

0 - Left-pads the number with zeroes (0) instead of spaces, where padding is specified.
4 (width) - Minimum number of characters to be printed. If the value to be printed is shorter than this number, the result is right justified within this width by padding on the left with the pad character. By default this is a blank space, but the leading zero we used specifies a zero as the pad char. The value is not truncated even if the result is larger.
x - Specifier for hexadecimal integer.

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Related Solutions

C Programming – Printing Hexadecimal Characters

You are seeing the ffffff because char is signed on your system. In C, vararg functions such as printf will promote all integers smaller than int to int. Since char is an integer (8-bit signed integer in your case), your chars are being promoted to int via sign-extension.

Since c0 and 80 have a leading 1-bit (and are negative as an 8-bit integer), they are being sign-extended while the others in your sample don't.

char    int
c0 -> ffffffc0
80 -> ffffff80
61 -> 00000061

Here's a solution:

char ch = 0xC0;
printf("%x", ch & 0xff);

This will mask out the upper bits and keep only the lower 8 bits that you want.

Best Answer

Related Solutions

C Programming – Printing Hexadecimal Characters

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