When the C/C++ language printf function outputs the value of the array pointed to by the pointer, why does the value of the pointer change like this?

Question

1. In what order are the parameters of the prinf function containing pointer expressions calculated? The result of the code operation is obviously not from left to right. {code...}

给我你的怀抱 · Answer

Changing a variable multiple times in one statement is undefined behavior and may have different results on different platforms. This question makes no sense.

学习ing · Answer

printf{"%d",++i} represents two operations

First execute i=i+1, then output i

And i++ means

First output, then execute i=i+1

某草草 · Answer

The order of operation of function parameters has little to do with the internal logic of the function. It should calculate ++p before pushing it onto the stack (before the function is executed), and then calculate p++ after the function ends. If you want to know the specific sequence, you can refer to the assembly code (the specific meaning will be updated tomorrow, sorry)

    .file    "a.cpp"
    .def    ___main;    .scl    2;    .type    32;    .endef
    .section .rdata,"dr"
LC0:
    .ascii "%d1
3 2 1 1"
LC1:
    .ascii "%d %d %d %drrreee"
    .text
    .globl    _main
    .def    _main;    .scl    2;    .type    32;    .endef
_main:
LFB10:
    .cfi_startproc
    pushl    %ebp
    .cfi_def_cfa_offset 8
    .cfi_offset 5, -8
    movl    %esp, %ebp
    .cfi_def_cfa_register 5
    pushl    %esi
    pushl    %ebx
    andl    $-16, %esp
    subl    , %esp
    .cfi_offset 6, -12
    .cfi_offset 3, -16
    call    ___main
    movl    , 40(%esp)
    movl    , 44(%esp)
    movl    , 48(%esp)
    movl    , 52(%esp)
    movl    , 56(%esp)
    leal    40(%esp), %eax
    movl    %eax, 60(%esp)
    movl    60(%esp), %eax
    movl    (%eax), %eax
    movl    %eax, 4(%esp)
    movl    $LC0, (%esp)
    call    _printf
    movl    60(%esp), %eax
    movl    (%eax), %ebx
    movl    60(%esp), %eax
    leal    4(%eax), %edx
    movl    %edx, 60(%esp)
    movl    (%eax), %ecx
    movl    60(%esp), %eax
    movl    (%eax), %edx
    addl    , 60(%esp)
    movl    60(%esp), %eax
    leal    4(%eax), %esi
    movl    %esi, 60(%esp)
    movl    (%eax), %eax
    movl    %ebx, 16(%esp)
    movl    %ecx, 12(%esp)
    movl    %edx, 8(%esp)
    movl    %eax, 4(%esp)
    movl    $LC1, (%esp)
    call    _printf
    call    ___getreent
    movl    4(%eax), %eax
    movl    %eax, (%esp)
    call    _getc
    movl    rrreee, %eax
    leal    -8(%ebp), %esp
    popl    %ebx
    .cfi_restore 3
    popl    %esi
    .cfi_restore 6
    popl    %ebp
    .cfi_restore 5
    .cfi_def_cfa 4, 4
    ret
    .cfi_endproc
LFE10:
    .ident    "GCC: (GNU) 5.4.0"
    .def    _printf;    .scl    2;    .type    32;    .endef
    .def    ___getreent;    .scl    2;    .type    32;    .endef
    .def    _getc;    .scl    2;    .type    32;    .endef

Digression:
2 and 3 that appear in the results can still be explained.
4 is very strange. If I have to explain it reluctantly, the ++ outside the brackets of *(++p)++ also works for p,
but the operator is in the form of p++, which should be in It is incremented after the statement ends, so this explanation is obviously wrong.
I am in Cygwin + gcc (GCC) 5.4.0environment. The running results are as follows. 4 does not appear. What environment did you use?

rrreee

ringa_lee · Answer

The order in which function parameters are pushed onto the stack is certain, but the order in which the parameters are evaluated is not specified. The
compiler only guarantees that the values of all parameters are known before printf is called
Information on this is available Search Sequence Point