From the glibc sprintf() documentation: The behavior of this function is undefined if copying takes place between objects that overlap—for example, if s is also given as an argument to be printed under control of the ‘%s’ conversion. It may be safe in a particular implementation; but you could not count on it being portable. … Read more
The trick is to use type classes. In the case of printf, the key is the PrintfType type class. It does not expose any methods, but the important part is in the types anyway. class PrintfType r printf :: PrintfType r => String -> r So printf has an overloaded return type. In the trivial … Read more
Pure Bash, no external utilities This demonstration does full justification, but you can just omit subtracting the length of the second string if you want ragged-right lines. pad=$(printf ‘%0.1s’ “-“{1..60}) padlength=40 string2=’bbbbbbb’ for string1 in a aa aaaa aaaaaaaa do printf ‘%s’ “$string1” printf ‘%*.*s’ 0 $((padlength – ${#string1} – ${#string2} )) “$pad” printf ‘%s\n’ … Read more
The # part gives you a 0x in the output string. The 0 and the x count against your “8” characters listed in the 08 part. You need to ask for 10 characters if you want it to be the same. int i = 7; printf(“%#010x\n”, i); // gives 0x00000007 printf(“0x%08x\n”, i); // gives 0x00000007 … Read more
Declare an extra variable before using printf: String format = “%” + fieldSize + “d”; System.out.printf(format, yourVariables); (this is the first solution I found from a web search)
The short answer is that it has no impact on printf, and denotes use of float or double in scanf. For printf, arguments of type float are promoted to double so both %f and %lf are used for double. For scanf, you should use %f for float and %lf for double. More detail for the … Read more
You can’t get the filename exactly as input; the shell will handle all that redirection stuff without telling you. In the case of a direct < file redirection, you can retrieve a filepath associated with stdin by using fstat to get an inode number for it then walking the file hierarchy similarly to find / … Read more
There’s probably no easier way. It’s a quite involved problem. Your code isn’t solving it right for several reasons: Most practical implementations of floating-point arithmetic aren’t decimal, they are binary. So, when you multiply a floating-point number by 10 or divide it by 10, you may lose precision (this depends on the number). Even though … Read more
The code has undefined behaviour. In Turbo C++, it just so happens that the three variables live at the exact positions on the stack where the missing printf() argument would be. This results in the undefined behaviour manifesting itself by having the “correct” values printed. However, you can’t reasonably rely on this to be the … Read more
fflush() flushes buffered output in line or full-buffered stdio streams: extern fflush … xor edi, edi ; RDI = 0 call fflush ; fflush(NULL) flushes all streams … Alternatively, mov rdi, [stdout] / call fflush also works to flush only that stream. (Use default rel for efficient RIP-relative addressing, and you’ll need extern stdout as … Read more