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I know that when using %x with printf() we are printing 4 bytes (an int in hexadecimal) from the stack. But I would like to print only 1 byte. Is there a way to do this ?
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Unsigned Integer Format Specifier %u The %u format specifier is implemented for fetching values from the address of a variable having an unsigned decimal integer stored in the memory. It is used within the printf() function for printing the unsigned integer variable.
The %a formatting specifier is new in C99. It prints the floating-point number in hexadecimal form. This is not something you would use to present numbers to users, but it's very handy for under-the-hood/technical use cases.
In C programming language, %d and %i are format specifiers as where %d specifies the type of variable as decimal and %i specifies the type as integer. In usage terms, there is no difference in printf() function output while printing a number using %d or %i but using scanf the difference occurs.
The Printf module API details the type conversion flags, among them: %B: convert a boolean argument to the string true or false %b: convert a boolean argument (deprecated; do not use in new programs).
Assumption:You want to print the value of a variable of 1 byte width, i.e., char.
In case you have a char variable say, char x = 0; and want to print the value, use %hhx format specifier with printf().
Something like
printf("%hhx", x);
Otherwise, due to default argument promotion, a statement like
printf("%x", x);
would also be correct, as printf() will not read the sizeof(unsigned int) from stack, the value of x will be read based on it's type and the it will be promoted to the required type, anyway.
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