4 ms·

Compilers didn't replace any jobs, they created more. Similarly, this type of AI-assisted programming will allow more people to program and make existing progra

by axg11 5y ago

Compilers didn't replace any jobs, they created more. Similarly, this type of AI-assisted programming will allow more people to program and make existing programmers more productive.

keewee7 5y ago
I was thinking over a really long time period. There is at least 20-30 more years of general purpose programming being a highly sought after skill. But with time most programming is going to be done by AI that is directed by domain experts.
jeffbee 5y ago
In my view this type of system will only be usable by Real Computer Scientists and will completely kill off the workaday hacker. Think of all the people who bitterly complain that a C++ compiler does something unexpected under the banner of UB. That crowd cannot cope with a world in which you have to exactly describe your requirements to an AI. It is also analogous to TDD, so all the TDD haters, which is the overwhelming majority of hackers, are toast.
- meken 5y ago
  What’s a workaday hacker and what’s UB?
  - 331c8c71 5y ago
    undefined behavior
  - kilotaras 5y ago
    UB = undefined behaviour. You can write code that is valid as in "can be compiled" but outside of C++ standard. It is duty of programmer to not have those, as compiler usually assumes that there's no UB in your code and can do unintuitive things with optimizations. e.g int foo(int8_t x) { x += 120 return x; } int bar(int8_t y) { int z = foo(y); if (y > 8) { do_important_thing(z); } } `do_important_thing` may be optimized out because: 1. signed overflow is a UB. Compiler than assumes that everything passed to foo is less than 8; 2. We pass y to foo => y < 8 3. if branch can then be optimized out
    - jcranmer 5y ago
      To be pedantic, C has no 8- or 16-bit addition operators, since everything sub-int is scaled up to int to do arithmetic. Therefore, the `x += 120;` line never overflows, since it is actually `x = (int8_t)((int)x + 120);`, and the possible range of `(int)x + 120` is comfortably within the range of expressible ints, while the conversion to int8_t is defined to wrap around when oversized. So there compiler can't optimize out do_important_thing in your example.
      - kilotaras 5y ago
        Huh, TIL. I was too lazy to check the exact value of 2^31 and payed for it.
      - jeffbee 5y ago
        This version of the future belongs to the pedants.
- discreteevent 5y ago
  I hope you don't mean to imply that TDD haters are not real computer scientists. http://ravimohan.blogspot.com/2007/04/learning-from-sudoku-solvers.html?m=1 http://ravimohan.blogspot.com/2007/04/learning-from-sudoku-s...

by axg11 5y ago

Compilers didn't replace any jobs, they created more. Similarly, this type of AI-assisted programming will allow more people to program and make existing programmers more productive.

keewee7 5y ago

I was thinking over a really long time period. There is at least 20-30 more years of general purpose programming being a highly sought after skill. But with time most programming is going to be done by AI that is directed by domain experts.

jeffbee 5y ago

In my view this type of system will only be usable by Real Computer Scientists and will completely kill off the workaday hacker. Think of all the people who bitterly complain that a C++ compiler does something unexpected under the banner of UB. That crowd cannot cope with a world in which you have to exactly describe your requirements to an AI. It is also analogous to TDD, so all the TDD haters, which is the overwhelming majority of hackers, are toast.

meken 5y ago
What’s a workaday hacker and what’s UB?
- 331c8c71 5y ago
  undefined behavior
- kilotaras 5y ago
  UB = undefined behaviour. You can write code that is valid as in "can be compiled" but outside of C++ standard. It is duty of programmer to not have those, as compiler usually assumes that there's no UB in your code and can do unintuitive things with optimizations. e.g int foo(int8_t x) { x += 120 return x; } int bar(int8_t y) { int z = foo(y); if (y > 8) { do_important_thing(z); } } `do_important_thing` may be optimized out because: 1. signed overflow is a UB. Compiler than assumes that everything passed to foo is less than 8; 2. We pass y to foo => y < 8 3. if branch can then be optimized out
  - jcranmer 5y ago
    To be pedantic, C has no 8- or 16-bit addition operators, since everything sub-int is scaled up to int to do arithmetic. Therefore, the `x += 120;` line never overflows, since it is actually `x = (int8_t)((int)x + 120);`, and the possible range of `(int)x + 120` is comfortably within the range of expressible ints, while the conversion to int8_t is defined to wrap around when oversized. So there compiler can't optimize out do_important_thing in your example.
    - kilotaras 5y ago
      Huh, TIL. I was too lazy to check the exact value of 2^31 and payed for it.
    - jeffbee 5y ago
      This version of the future belongs to the pedants.
discreteevent 5y ago
I hope you don't mean to imply that TDD haters are not real computer scientists. http://ravimohan.blogspot.com/2007/04/learning-from-sudoku-solvers.html?m=1 http://ravimohan.blogspot.com/2007/04/learning-from-sudoku-s...