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Calculate number of days from unix-timestamp in math?

How can I calculate the day number from unix-timestamp mathematically and without using any functions and in a simple mathematical formula.

1313905026 → 8 (Today 08/21/2011)

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3 answers

There is no simple formula for this. You will need to subtract the number of years (given leap years) from the era, which will probably require a loop or discrete calculation. Then use some type of cycle to subtract the number of seconds in each month for the current year. Only a few seconds are left at the moment.

I would do something like this.

x = ...//the number of seconds
year = 1970

while (x > /*one year*/){
 x = x - /*seconds in january, and march-december*/
 if(year % 4 == 0){
  x -= /*leapeay seconds in february*/
 }else{
  x -= /*regular seconds in february*/
 }
}

//Then something like this:

if(x > /*seconds in january*/){
 x -= /*seconds in january*/
}
if(x > /*seconds in february*/){
 x -= /*seconds in january*/
}

.
.
.

//After that just get the number of days from x seconds and you're set.

Edit

, , , , - , .

t - .

F - . . : 126230400.

, , F, : y.

, y = n% F.

: 1. y , 2. y 3. y 4. y 5. y

, 1972 , , 1970 , , .

let jan, feb, febLY, mar, may,..., dec - ( ).

d - , D - (86400). y , yLY - .

y = (t % F)
if(y < Y){
 if(y > jan){
  y -= jan
 }
 if(y > feb){
  y -= feb
 }
 .
 .
 .
 d = y % D
}
else if(y < 2 * y){
 y = y - Y
 if(y > jan){
  y -= jan
 }
 if(y > feb){
  y -= feb
 }
 .
 .
 .
 d = y % D
}
else if(y < 2 * y + yLY){
 y = y - 2 * Y
 if(y > jan){
  y -= jan
 }
 if(y > febLY){
  y -= febLY
 }
 .
 .
 .
 d = y % D
}
else{
 y = y - 2 * Y - yLY
 if(y > jan){
  y -= jan
 }
 if(y > feb){
  y -= feb
 }
 .
 .
 .
 d = y % D
}

. , 1 /24 , . , .

+3

A unix timestamp , . 1 y/m/d unix timestamp:

#include <iostream>

int
main()
{
    int s = 1313905026;
    int z = s / 86400 + 719468;
    int era = (z >= 0 ? z : z - 146096) / 146097;
    unsigned doe = static_cast<unsigned>(z - era * 146097);
    unsigned yoe = (doe - doe/1460 + doe/36524 - doe/146096) / 365;
    int y = static_cast<int>(yoe) + era * 400;
    unsigned doy = doe - (365*yoe + yoe/4 - yoe/100);
    unsigned mp = (5*doy + 2)/153;
    unsigned d = doy - (153*mp+2)/5 + 1;
    unsigned m = mp + (mp < 10 ? 3 : -9);
    y += (m <= 2);
    std::cout << m << '/' << d << '/' << y << '\n'; // 8/21/2011
}

:

8/21/2011

y m ( d), .

. , ( ).

1 : , , clang -O3 macOS:

__Z14get_day_numberi:                   ## @_Z14get_day_numberi
    .cfi_startproc
## BB#0:
    pushq   %rbp
Ltmp0:
    .cfi_def_cfa_offset 16
Ltmp1:
    .cfi_offset %rbp, -16
    movq    %rsp, %rbp
Ltmp2:
    .cfi_def_cfa_register %rbp
    movslq  %edi, %rax
    imulq   $-1037155065, %rax, %rcx ## imm = 0xFFFFFFFFC22E4507
    shrq    $32, %rcx
    addl    %ecx, %eax
    movl    %eax, %ecx
    shrl    $31, %ecx
    sarl    $16, %eax
    leal    (%rax,%rcx), %edx
    leal    719468(%rax,%rcx), %esi
    testl   %esi, %esi
    leal    573372(%rax,%rcx), %eax
    cmovnsl %esi, %eax
    cltq
    imulq   $963315389, %rax, %rcx  ## imm = 0x396B06BD
    movq    %rcx, %rsi
    shrq    $63, %rsi
    shrq    $32, %rcx
    sarl    $15, %ecx
    addl    %esi, %ecx
    imull   $146097, %ecx, %ecx     ## imm = 0x23AB1
    movl    %eax, %esi
    subl    %ecx, %esi
    subl    %eax, %esi
    leal    719468(%rsi,%rdx), %eax
    movl    %eax, %ecx
    shrl    $2, %ecx
    imulq   $1506180313, %rcx, %rdx ## imm = 0x59C67CD9
    shrq    $39, %rdx
    movl    %eax, %esi
    subl    %edx, %esi
    imulq   $963321983, %rcx, %rcx  ## imm = 0x396B207F
    shrq    $43, %rcx
    addl    %esi, %ecx
    movl    %eax, %edx
    shrl    $4, %edx
    imulq   $7525953, %rdx, %rdx    ## imm = 0x72D641
    shrq    $36, %rdx
    subl    %edx, %ecx
    imulq   $1729753953, %rcx, %rsi ## imm = 0x6719F361
    shrq    $32, %rsi
    movl    %ecx, %r8d
    subl    %ecx, %eax
    movl    %ecx, %edi
    movl    $3855821599, %edx       ## imm = 0xE5D32B1F
    imulq   %rcx, %rdx
    subl    %esi, %ecx
    shrl    %ecx
    addl    %esi, %ecx
    shrl    $8, %ecx
    imull   $365, %ecx, %ecx        ## imm = 0x16D
    subl    %ecx, %r8d
    shrl    $2, %edi
    imulq   $1506180313, %rdi, %rcx ## imm = 0x59C67CD9
    shrq    $39, %rcx
    shrq    $47, %rdx
    addl    %r8d, %eax
    subl    %ecx, %eax
    leal    (%rax,%rdx), %ecx
    leal    2(%rcx,%rcx,4), %esi
    movl    $3593175255, %edi       ## imm = 0xD62B80D7
    imulq   %rsi, %rdi
    shrq    $39, %rdi
    imull   $153, %edi, %edi
    subl    %edi, %esi
    leal    4(%rcx,%rcx,4), %ecx
    subl    %esi, %ecx
    movl    $3435973837, %esi       ## imm = 0xCCCCCCCD
    imulq   %rcx, %rsi
    shrq    $34, %rsi
    leal    1(%rax,%rdx), %eax
    subl    %esi, %eax
    popq    %rbp
    retq
    .cfi_endproc
+4
t = unix time
second = t MOD 60  
minute = INT(t / 60) MOD 60  
hour = INT(t / 60 / 60) MOD 24  
days = INT(t / 60 / 60 / 24)  
years = INT(days / 365.25)  
year = 1970 + years + 1

1970 , :

weekday = (days + 4) MOD 7

- 0. , ​​1, 1.

, .

days = days - years * 365 - leapdays

, .

IF year MOD 4 = 0 THEN ly = 1 ELSE ly = 0
WHILE month <= 12
    month = month + 1
    IF month = 2 THEN
        DaysInMonth = 28 + NOT(year MOD 4) + NOT(year MOD 100)
            + NOT(year MOD 400)
    ELSE
        DaysInMonth = 30 + (month + (month < 7)) MOD 2
    END IF
    IF days > DaysInMonth THEN days = days - DaysInMonth
END WHILE

This assumes Boolean values ​​TRUE = 1, FALSE = 0, NOT TRUE = 0, and NOT FALSE = 1.

Now we have the year, month, day of the month, hour, minute and second, calculated with adjustments for leap years.

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