printf

问题 #include <stdio.h> struct mychar { char value; struct mychar *nextPtr; }; typedef struct mychar Mychar; void instructions(); void append(Mychar **, char ); void printlist(Mychar *); int main(){ instructions(); Mychar *startPtr = NULL; unsigned int choice; char newchar; do { scanf("%d",&choice); switch (choice) { case 1: printf("\nWrite the character you want to add."); printf("\n> "); scanf(" %c", &newchar); append(&startPtr, newchar); printlist(startPtr); break; case 2: break; default:

问题 #include <stdio.h> struct mychar { char value; struct mychar *nextPtr; }; typedef struct mychar Mychar; void instructions(); void append(Mychar **, char ); void printlist(Mychar *); int main(){ instructions(); Mychar *startPtr = NULL; unsigned int choice; char newchar; do { scanf("%d",&choice); switch (choice) { case 1: printf("\nWrite the character you want to add."); printf("\n> "); scanf(" %c", &newchar); append(&startPtr, newchar); printlist(startPtr); break; case 2: break; default:

问题 #include <stdio.h> struct mychar { char value; struct mychar *nextPtr; }; typedef struct mychar Mychar; void instructions(); void append(Mychar **, char ); void printlist(Mychar *); int main(){ instructions(); Mychar *startPtr = NULL; unsigned int choice; char newchar; do { scanf("%d",&choice); switch (choice) { case 1: printf("\nWrite the character you want to add."); printf("\n> "); scanf(" %c", &newchar); append(&startPtr, newchar); printlist(startPtr); break; case 2: break; default:

问题 int x = 0xff; printf("%#x",x); Output: 0xff printf("%x",x); Ouput: ff Why is there a difference in output? What does # specifically do? 回答1: The standard says: 7.21.6 - 2 The result is converted to an ‘‘alternative form’’. ... For x (or X) conversion, a nonzero result has 0x (or 0X) prefixed to it. It does other interesting stuff (especially for floats) but I have rarely seen it used - I honestly admit I had to look it up to remember. 来源： https://stackoverflow.com/questions/11922876/what-does

问题 I have a program that counts root of quadratic equation. And I have a problem with printing the results on the screen, because I can print only one value. This is my code below, could you please tell me what should I do to pass two result to expression "x1 = ... ". [bits 32] call getaddr format db "x1 = %lf, x2 = %lf", 0xA, 0 offset equ $ - format a dq 1.0 ; b dq -11.0 c dq 28.0 minusfour dq -4.0 getaddr: finit mov eax, [esp] lea eax, [eax+offset] ; eax = a mov edx, [esp] lea edx, [edx+offset

问题 Can I pass a negative int in printf while printing through format specifier %c since while printing int gets converted into an unsigned char? Is printf("%c", -65); valid? — I tried it on GCC but getting a diamond-like character(with question-mark inside) as output. Why? 回答1: Absolutely yes, if char is a signed type. C allows char to be signed or unsigned and in GCC you can switch between them with -funsigned-char and -fsigned-char. When char is signed it's exactly the same thing as this char

问题 We can print the current time with the builtin printf function, without needing to invoke an external command like date , like this: printf '%(%Y-%m-%d:%H:%M:%S)T %s\n' -1 # sample output: 2019-03-30:17:39:36,846 How can we make printf to print milliseconds or nanoseconds as well? Using %3N or %N in the format string doesn't work: printf '%(%Y-%m-%d:%H:%M:%S,%3N)T %s\n' -1 # outputs 2019-03-30:17:38:16,%3N printf '%(%Y-%m-%d:%H:%M:%S,%N)T %s\n' -1 # outputs 2019-03-30:17:38:16,%N However, the

问题 We can print the current time with the builtin printf function, without needing to invoke an external command like date , like this: printf '%(%Y-%m-%d:%H:%M:%S)T %s\n' -1 # sample output: 2019-03-30:17:39:36,846 How can we make printf to print milliseconds or nanoseconds as well? Using %3N or %N in the format string doesn't work: printf '%(%Y-%m-%d:%H:%M:%S,%3N)T %s\n' -1 # outputs 2019-03-30:17:38:16,%3N printf '%(%Y-%m-%d:%H:%M:%S,%N)T %s\n' -1 # outputs 2019-03-30:17:38:16,%N However, the

问题 I'm trying to write a program in C++ which will be responsible for simulating blinkers in cars. I want it to be simple and to compile it in a console window. Is it possible to create one thread for input which will be always active and second for output that will run simultaneously? I wanted to use threads to solve this but it doesn't work as I would like. I have a little trouble to understand threads. If anyone could help me to fix this I would be grateful. int in() { int i; cout<<"press 1

问题 I want to print several arrays and the element of output will with field width 3 , I think I can use printf , but if I use printf then I need to write the format of all element of array , but the array is big . for example @array = (1,10,100,30); printf ("%3d %3d %3d %3d\n",$array[0],$array[1],$array[2],$array[3]); I know I can use loop to print a element until all the array loop through , but I think it's not a good idea . Does there exists any way can let me just describe the format of