If you have your “09” in a variable, you can do a=”09″ echo “$a” echo “${a#0}” printf “%04d” “${a#0}” Why does this help? Well, a number literal starting with 0 but having no x at the 2nd place is interpreted as octal value. Octal value only have the digits 0..7, 8 and 9 are unknown. … Read more
You’ve done it. this is argument 1, so by saying format(printf, 2, 3) you’re telling the compiler that you’re NOT printing this, you’re printing argument 2 (fmt) with additional arguments past that.
It looks like the Right Way to do this in F# 2.0 is by using the StructuredFormatDisplay attribute, for example: [<StructuredFormatDisplay(“hello {a}”)>] type myType = {a: int} In this example, instead of the default {a = 42;}, you would get hello 42. This works the same way for object, record, and union types. And although … Read more
That would be the following macros from inttypes.h: For printf: PRIdPTR PRIiPTR PRIoPTR PRIuPTR PRIxPTR PRIXPTR For scanf: SCNdPTR SCNiPTR SCNoPTR SCNuPTR SCNxPTR Usage example: uintptr_t p = SOME_VALUE; printf(“Here’s a pointer for you: %” PRIxPTR “\n”, p);
Yes, float arguments to variadic function are promoted to double. The draft C99 standard section 6.5.2.2 Function calls says: […]and arguments that have type float are promoted to double. These are called the default argument promotions.[…] from the draft C++ standard section 5.2.2 Function call: […]a floating point type that is subject to the floating … Read more
Online C Draft Standard (n1256), section 7.19.6.1, paragraph 2: The fprintf function writes output to the stream pointed to by stream, under control of the string pointed to by format that specifies how subsequent arguments are converted for output. If there are insufficient arguments for the format, the behavior is undefined. If the format is … Read more
You’re probably getting a benign form of undefined behaviour because the %x modifier expects an unsigned int parameter and a char will usually be promoted to an int when passed to a varargs function. You should explicitly cast the char to an unsigned int to get predictable results: printf(” 0x%1x “, (unsigned)pixel_data[0] ); Note that … Read more
Output is buffered. stdout is line-buffered by default, which means that ‘\n’ is supposed to flush the buffer. Why is it not happening in your case? I don’t know. I need more info about your application/environment. However, you can control buffering with setvbuf(): setvbuf(stdout, NULL, _IOLBF, 0); This will force stdout to be line-buffered. setvbuf(stdout, … Read more
Your problem is that you have to change: printf(“%”); to printf(“%%”); Or you could use ASCII code and write: printf(“%c”, 37); 🙂
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. … Read more