Simply cast your int pointer to a void one: printf( “Address of p1: %p\n”, ( void * )pt1 ); Your code is safe, but you are compiling with the -Wformat warning flag, that will type check the calls to printf() and scanf().
The width specifier is the complete width: printf(“%05.1f\n”, myVar); // Total width 5, pad with 0, one digit after . To get your expected format: printf(“% 06.1f\n”, myVar);
Because conversion of wide characters is done according to the currently set locale. By default a C program always starts with a “C” locale which only supports ASCII characters. You have to switch to any Russian or UTF-8 locale first: setlocale(LC_ALL, “ru_RU.utf8”); // Russian Unicode setlocale(LC_ALL, “en_US.utf8”); // English US Unicode Or to a current … Read more
There seems to be no perfect way. The root of the problem is that clock_t can be either integer or floating point. clock_t can be a floating point type As Bastien Léonard mentions for POSIX (go upvote him), C99 N1256 draft 7.23.1/3 also says that: [clock_t is] arithmetic types capable of representing times and 6.2.5/18: … Read more
Yes, if you use * in your format string, it gets a number from the arguments: printf (“%0*d\n”, 3, 5); will print “005”. Keep in mind you can only pad with spaces or zeros. If you want to pad with something else, you can use something like: #include <stdio.h> #include <string.h> int main (void) { … Read more
You can use the printf command, like this: printf “%-10s |\n” “$txt” The %s means to interpret the argument as string, and the -10 tells it to left justify to width 10 (negative numbers mean left justify while positive numbers justify to the right). The \n is required to print a newline, since printf doesn’t … Read more
The character sequence \012 inside the string is interpreted as an octal escape sequence. The value 012 interpreted as octal is 10 in decimal, which is the line feed (\n) character on most terminals. From the Wikipedia page: An octal escape sequence consists of \ followed by one, two, or three octal digits. The octal … Read more
I believe it is technically undefined, because the “correct type” for %x is specified as unsigned int – and as you point out, there is no exception for signed/unsigned mismatch here. The rules for printf are for a more specific case and thus override the rules for the general case (for another example of the … Read more
There are a couple of approaches. You could write your code in C99-conforming fashion, and then supply system-specific hacks when the compiler-writers let you down. (Sadly, that’s rather common in C99.) #include <stdint.h> #include <inttypes.h> printf(“My value is %10” PRId64 “\n”, some_64_bit_expression); If one of your target systems has neglected to implement <inttypes.h> or has … Read more
The backtick operator removes trailing new lines. See 3.4.5. Command substitution at http://tldp.org/LDP/Bash-Beginners-Guide/html/sect_03_04.html Note on edited question Compare: [alvaro@localhost ~]$ printf “\n” [alvaro@localhost ~]$ echo “\n” \n [alvaro@localhost ~]$ echo -e “\n” [alvaro@localhost ~]$ The echo command doesn’t treat \n as a newline unless you tell him to do so: NAME echo – display a … Read more