You reassigned *a on the first line, and it's using that new value on the second line. You'll need a temporary to avoid that, e.g.:

int origa = *a;
*a = abs(origa + *b);
*b = abs(origa - *b);

Alternatively, you can get fancy with tuples:

std::tie(*a, *b) = std::make_tuple(abs(*a + *b), abs(*a - *b));

which just computes and packs up the values first, then unpacks them after all the reads are done.

Alright, the following works. @user16217248 got me started. See the discussion under that answer.

How to safely and efficiently find abs((int)num1 - (int)num2)

/// Safely and efficiently return `abs((int)num1 - (int)num2)`
unsigned int abs_num1_minus_num2_int(int num1, int num2)
{
    unsigned int abs_diff = num1 > num2 ?
        (unsigned int)num1 - (unsigned int)num2 :
        (unsigned int)num2 - (unsigned int)num1;

    return abs_diff;
}

The secret is to do the num1 > num2 ternary comparison with signed integer values, but then to reinterpret cast them to unsigned integer values to allow for well-defined overflow and underflow behavior when getting the absolute value of num1 - num2.

Here is my full test code:

absolute_value_of_num1_minus_num2.c from my eRCaGuy_hello_world repo:

///usr/bin/env ccache gcc -Wall -Wextra -Werror -O3 -std=gnu17 "$0" -o /tmp/a -lm && /tmp/a "$@"; exit
// For the line just above, see my answer here: https://stackoverflow.com/a/75491834/4561887

#include <limits.h>
#include <stdbool.h> // For `true` (`1`) and `false` (`0`) macros in C
#include <stdint.h>  // For `uint8_t`, `int8_t`, etc.
#include <stdio.h>   // For `printf()`


#define TEST_EQ(func, num1, num2, equals) \
    printf("%s\n", func((num1), (num2)) == (equals) ? "Passed" : "FAILED")

/// Safely and efficiently return `abs((int8_t)num1 - (int8_t)num2)`
uint8_t abs_num1_minus_num2_int8(int8_t num1, int8_t num2)
{
    // Note: due to implicit type promotion rules, rule 2 in my answer here
    // (https://stackoverflow.com/a/72654668/4561887) applies, and both the `>`
    // comparison, as well as subtraction, take place below as `int` types.
    // While signed `int` overflow and underflow is undefined behavior, none of
    // that occurs here.
    // - It's just useful to understand that even though we are doing
    //   `(uint8_t)num1 -(uint8_t)num2`, the C compiler really sees it as this:
    //   `(int)(uint8_t)num1 - (int)(uint8_t)num2`.
    // - This is because all small types smaller than `int`, such as `uint8_t`,
    //   are first automatically implicitly cast to `int` before any
    //   mathematical operation or comparison occurs.
    // - The C++ compiler therefore sees it as this:
    //   `static_cast<int>(static_cast<unsigned char>(num1)) - static_cast<int>(static_cast<unsigned char>(num2))`.
    //   - Run this code through https://cppinsights.io/ to see that.
    //     See here: https://cppinsights.io/s/bfc425f6 --> and click the play
    //     button.
    uint8_t abs_diff = num1 > num2 ?
        (uint8_t)num1 - (uint8_t)num2 :
        (uint8_t)num2 - (uint8_t)num1;

    // debugging
    printf("num1 = %4i (%3u); num2 = %4i (%3u); num1-num2=%3u;  ",
        num1, (uint8_t)num1, num2, (uint8_t)num2, abs_diff);

    return abs_diff;
}

/// Safely and efficiently return `abs((int)num1 - (int)num2)`
unsigned int abs_num1_minus_num2_int(int num1, int num2)
{
    unsigned int abs_diff = num1 > num2 ?
        (unsigned int)num1 - (unsigned int)num2 :
        (unsigned int)num2 - (unsigned int)num1;

    // debugging
    printf("num1 = %11i (%10u); num2 = %11i (%10u); num1-num2=%10u;  ",
        num1, (unsigned int)num1, num2, (unsigned int)num2, abs_diff);

    return abs_diff;
}


int main()
{
    printf("Absolute difference tests.\n");

    // ---------------
    // int8_t types
    // ---------------

    int8_t num1_8;
    int8_t num2_8;

    printf("\n");
    printf("INT8_MIN = %i, INT8_MAX = %i\n", INT8_MIN, INT8_MAX); // -128, 127

    num1_8 = -7;
    num2_8 = -4;
    TEST_EQ(abs_num1_minus_num2_int8, num1_8, num2_8, 3);

    num1_8 = INT8_MIN;
    num2_8 = INT8_MAX;
    TEST_EQ(abs_num1_minus_num2_int8, num1_8, num2_8, UINT8_MAX);

    num1_8 = INT8_MAX;
    num2_8 = INT8_MIN;
    TEST_EQ(abs_num1_minus_num2_int8, num1_8, num2_8, UINT8_MAX);

    num1_8 = 100;
    num2_8 = 10;
    TEST_EQ(abs_num1_minus_num2_int8, num1_8, num2_8, 90);

    num1_8 = 10;
    num2_8 = 100;
    TEST_EQ(abs_num1_minus_num2_int8, num1_8, num2_8, 90);

    num1_8 = 10;
    num2_8 = 10;
    TEST_EQ(abs_num1_minus_num2_int8, num1_8, num2_8, 0);

    num1_8 = INT8_MAX;
    num2_8 = 1;
    TEST_EQ(abs_num1_minus_num2_int8, num1_8, num2_8, 126);

    num1_8 = 1;
    num2_8 = INT8_MAX;
    TEST_EQ(abs_num1_minus_num2_int8, num1_8, num2_8, 126);

    // ---------------
    // int types
    // ---------------

    int num1;
    int num2;

    printf("\n");
    printf("INT_MIN = %i, INT_MAX = %i\n", INT_MIN, INT_MAX); // -2147483648, 2147483647

    num1 = -7;
    num2 = -4;
    TEST_EQ(abs_num1_minus_num2_int, num1, num2, 3);

    num1 = INT_MIN;
    num2 = INT_MAX;
    TEST_EQ(abs_num1_minus_num2_int, num1, num2, UINT_MAX);

    num1 = INT_MAX;
    num2 = INT_MIN;
    TEST_EQ(abs_num1_minus_num2_int, num1, num2, UINT_MAX);

    num1 = 100;
    num2 = 10;
    TEST_EQ(abs_num1_minus_num2_int, num1, num2, 90);

    num1 = 10;
    num2 = 100;
    TEST_EQ(abs_num1_minus_num2_int, num1, num2, 90);

    num1 = 10;
    num2 = 10;
    TEST_EQ(abs_num1_minus_num2_int, num1, num2, 0);

    num1 = INT_MAX;
    num2 = 1;
    TEST_EQ(abs_num1_minus_num2_int, num1, num2, 2147483646);

    num1 = 1;
    num2 = INT_MAX;
    TEST_EQ(abs_num1_minus_num2_int, num1, num2, 2147483646);


    return 0;
}

Sample run and output:

eRCaGuy_hello_world/c$ ./absolute_value_of_num1_minus_num2.c
Absolute difference tests.

INT8_MIN = -128, INT8_MAX = 127
num1 =   -7 (249); num2 =   -4 (252); num1-num2=  3;  Passed
num1 = -128 (128); num2 =  127 (127); num1-num2=255;  Passed
num1 =  127 (127); num2 = -128 (128); num1-num2=255;  Passed
num1 =  100 (100); num2 =   10 ( 10); num1-num2= 90;  Passed
num1 =   10 ( 10); num2 =  100 (100); num1-num2= 90;  Passed
num1 =   10 ( 10); num2 =   10 ( 10); num1-num2=  0;  Passed
num1 =  127 (127); num2 =    1 (  1); num1-num2=126;  Passed
num1 =    1 (  1); num2 =  127 (127); num1-num2=126;  Passed

INT_MIN = -2147483648, INT_MAX = 2147483647
num1 =          -7 (4294967289); num2 =          -4 (4294967292); num1-num2=         3;  Passed
num1 = -2147483648 (2147483648); num2 =  2147483647 (2147483647); num1-num2=4294967295;  Passed
num1 =  2147483647 (2147483647); num2 = -2147483648 (2147483648); num1-num2=4294967295;  Passed
num1 =         100 (       100); num2 =          10 (        10); num1-num2=        90;  Passed
num1 =          10 (        10); num2 =         100 (       100); num1-num2=        90;  Passed
num1 =          10 (        10); num2 =          10 (        10); num1-num2=         0;  Passed
num1 =  2147483647 (2147483647); num2 =           1 (         1); num1-num2=2147483646;  Passed
num1 =           1 (         1); num2 =  2147483647 (2147483647); num1-num2=2147483646;  Passed

Adjacently related

1. The "absolute subtraction" above reminds me of the "rounding divide" set of solutions you can do with integer math too. For that, see my other answer here: Rounding integer division (instead of truncating). I present rounding up, rounding down, and rounding to nearest when doing integer division.

See also

1. My answer on implicit casting/promotion and Integer and floating point rank and promotion rules in C and C++
2. https://cppinsights.io/ - a very useful tool which expands your C++ code into exactly what the compiler sees, including after applying all automatic implicit type promotion rules in the compiler.
   1. Ex: see my code above here: https://cppinsights.io/s/bfc425f6 --> then click the play button to convert and expand it into what the compiler sees.

The shortest solution in your first piece of code is to change the printf statement as follows:

    printf("absValue = %u\n", (unsigned)((u<0)?-u:u));

This will print the absolute value of u. The type conversion (unsigned) ensures that the data type is as expected by printf. The statement (u<0)?-u:u uses the conditional operator to select the value -u if the condition (u<0) is true and u if the condition is false (i.e. u>=0).

The problem in your code is that u is a signed integer which means its value is stored using the Two's complement representation in 4 bytes(*) and printf is not intelligent. When you tell printf to display an unsigned integer, then printf will take the 4 bytes holding u and interpret them as an unsigned integer. Since negative numbers in Two's complement are stored as large positive integers, that is the result you see.

(*) The use of Two's complement and the int size of 4 is machine-dependent, but common.

One common place where pointers are helpful is when you are writing functions. Functions take their arguments 'by value', which means that they get a copy of what is passed in and if a function assigns a new value to one of its arguments that will not affect the caller. This means that you couldn't write a "doubling" function like this:

void doubling(int x)
{
    x = x * 2;
}

This makes sense because otherwise what would the program do if you called doubling like this:

doubling(5);

Pointers provide a tool for solving this problem because they let you write functions that take the address of a variable, for example:

void doubling2(int *x)
{
    (*x) = (*x) * 2; 
}

The function above takes the address of an integer as its argument. The one line in the function body dereferences that address twice: on the left-hand side of the equal sign we are storing into that address and on the right-hand side we are getting the integer value from that address and then multiply it by 2. The end result is that the value found at that address is now doubled.

As an aside, when we want to call this new function we can't pass in a literal value (e.g. doubling2(5)) as it won't compile because we are not properly giving the function an address. One way to give it an address would look like this:

int a = 5;
doubling2(&a);

The end result of this would be that our variable a would contain 10.

The standard C library is providing the optimized solutions for many problems with considerations based on the architecture, compiler in use and others. The abs() function defined in stdlib.h is one of these, and it is used for your purpose exactly. To emphasize the point, here is ARM compiler result when using abs vs a version of a homebrew abs: https://arm.godbolt.org/z/aO7t1n

Paste:

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
    srand(111);
    int x = rand() - 200;
    printf("%d\n", abs(x));
}

results in

main:
        push    {r4, lr}
        mov     r0, #111
        bl      srand
        bl      rand
        sub     r1, r0, #200
        cmp     r1, #0
        rsblt   r1, r1, #0
        ldr     r0, .L4
        bl      printf
        mov     r0, #0
        pop     {r4, pc}
.L4:
        .word   .LC0
.LC0:
        .ascii  "%d\012\000"

And

#include <stdio.h>
#include <stdlib.h>

int my_abs(int x)
{
    return x < 0 ? -x : x;
}

int main(void)
{
    srand(111);
    int x = rand() - 200;
    printf("%d\n", my_abs(x));
}

results in

my_abs:
        cmp     r0, #0
        rsblt   r0, r0, #0
        bx      lr
main:
        push    {r4, lr}
        mov     r0, #111
        bl      srand
        bl      rand
        sub     r1, r0, #200
        cmp     r1, #0
        rsblt   r1, r1, #0
        ldr     r0, .L5
        bl      printf
        mov     r0, #0
        pop     {r4, pc}
.L5:
        .word   .LC0
.LC0:
        .ascii  "%d\012\000"

Notice that the main code is identical (only a label name is different) in both programs as my_abs got inlined, and its implementation is the same as the standard one.