integer

问题 What is %*d ? I know that %d is used for integers , so I think %*d also must related to integer only? What is the purpose of it? What does it do? int a=10,b=20; printf("\n%d%d",a,b); printf("\n%*d%*d",a,b); Result is 10 20 1775 1775 回答1: The %*d in a printf allows you to use a variable to control the field width, along the lines of: int wid = 4; printf ("%*d\n", wid, 42); which will give you: ..42 (with each of those . characters being a space). The * consumes one argument wid and the d

问题 I'm fairly new to programming; I've only been studying Python for a few weeks. I've been given an exercise recently that asks me to generate a list of integers, and then manually sort the numbers from lowest to highest in a separate list. import random unordered = list(range(10)) ordered = [] lowest = 0 i = 0 random.shuffle(unordered) lowest = unordered[0] while i in unordered: if unordered[i] < lowest: lowest = unordered[i] i += 1 if i >= len(unordered): i = 0 ordered.append(lowest)

问题 I'm fairly new to programming; I've only been studying Python for a few weeks. I've been given an exercise recently that asks me to generate a list of integers, and then manually sort the numbers from lowest to highest in a separate list. import random unordered = list(range(10)) ordered = [] lowest = 0 i = 0 random.shuffle(unordered) lowest = unordered[0] while i in unordered: if unordered[i] < lowest: lowest = unordered[i] i += 1 if i >= len(unordered): i = 0 ordered.append(lowest)

问题 The following program reverses user input. However, for numbers with trailing zeroes, the zeroes are 'ignored' when printing the reversed number. #include<stdio.h> int main(void) { int n, reversedNumber = 0, remainder; printf("Enter an integer: "); scanf("%d", &n); while(n != 0) { remainder = n%10; reversedNumber = reversedNumber*10 + remainder; n /= 10; } printf("Reversed Number = %d\n", reversedNumber); return 0; } Since the integer length of the user input is unknown , how can we print all

问题 The following program reverses user input. However, for numbers with trailing zeroes, the zeroes are 'ignored' when printing the reversed number. #include<stdio.h> int main(void) { int n, reversedNumber = 0, remainder; printf("Enter an integer: "); scanf("%d", &n); while(n != 0) { remainder = n%10; reversedNumber = reversedNumber*10 + remainder; n /= 10; } printf("Reversed Number = %d\n", reversedNumber); return 0; } Since the integer length of the user input is unknown , how can we print all

问题 hi i am having some problem with integer over flow in fortran 90. The maximum range of integer number is 10 digits.but, I want to deal with integer of the range 1E13. So what is the way of avoiding this? 回答1: define your variable as integer(kind=8) :: intVar To have a portable code, use the function selected_int_kind. In your case integer, parameter, k14 = selected_int_kind(14) integer(kind=k14) :: intVar selected_int_kind(r) returns the kind value of the smallest integer type that can

问题 hi i am having some problem with integer over flow in fortran 90. The maximum range of integer number is 10 digits.but, I want to deal with integer of the range 1E13. So what is the way of avoiding this? 回答1: define your variable as integer(kind=8) :: intVar To have a portable code, use the function selected_int_kind. In your case integer, parameter, k14 = selected_int_kind(14) integer(kind=k14) :: intVar selected_int_kind(r) returns the kind value of the smallest integer type that can

问题 hi i am having some problem with integer over flow in fortran 90. The maximum range of integer number is 10 digits.but, I want to deal with integer of the range 1E13. So what is the way of avoiding this? 回答1: define your variable as integer(kind=8) :: intVar To have a portable code, use the function selected_int_kind. In your case integer, parameter, k14 = selected_int_kind(14) integer(kind=k14) :: intVar selected_int_kind(r) returns the kind value of the smallest integer type that can

问题 I am using pinescript, and I have been trying to figure out why the following code does not work. The console keeps showing that series[integer] cannot output integer. I understand that series is not compatible with non-series values. If this is the case, is there a way to change series[integer] to integer? The following code does not work: x = barssince(crossover(cci,100)) y = barssince(crossover(100,cci)) xy = x-y //in this case the xy value is 9 z = highest(cci, abs(xy)) plot(z) The

问题 Suppose I have int n=123456; int x,y=0; How do I split the integer "n" in two half. Note : The Total Number of digits in n will always be multiple of 2, e.g. 1234, 4567, 234567, 345621 etc... all have 2,4,6,8 digits. I want to divide them in half. I am trying with following Code but it's not working, the y variable is holding reversed second part somehow. int x, y=0, len, digit; int n=123456; len=floor(log10(abs(n))) + 1; x=n; while((floor(log10(abs(x))) + 1)>len/2) { digit=x%10; x=x/10; y=(y