4 ms·

C is "portable assembler". If I do something like if (foo = bar) { } I know that the computer is comparing one value to another(witch means an op, eg.substra

C is "portable assembler". If I do something like if (foo = bar) { } I know that the computer is comparing one value to another(witch means an op, eg.substraction ), then doing a jump. I don't need to code the assembler, but I could estimate how much it takes everything really well(orders of magnitude). This is invaluable with coding. If you do: [object foo] in objective c or object.foo() You lost all you control, sometimes classes will use iterative methods or alloc and free memory every time you call a method, from nested methods, making it super slow. Sometimes I want abstraction, sometimes I want to know what is happening.

Xurinos 15y ago
This just isn't true. C is not portable assembler. It was never intended to be. I hear it claimed, and it is wrong every time somebody calls C a low-level language close to assembler. You can make some roughly reasonable assumptions about what comes out of the compiler, but often it is not what you think it is. Let's challenge this specific claim, that when you do "if (foo == bar)" -- I corrected the syntax error, which is a symptom of C's high-level syntax and not of the underlying assembly code -- you compare one value to another and then jump. For this challenge,I will write some trivial code that we should be able to make easy assumptions about, and I will compile it with debugging enabled so that I can dump the results with gdb. $ gcc -g example.c 1 #include <stdio.h> 2 3 int main() { 4 int foo = 10; 5 int bar = 20; 6 if (foo == bar) { 7 printf("Fun\n"); 8 } 9 return 0; 10 } Dump of assembler code for function main: 0x0000000100000ef8 <main+0>: push rbp 0x0000000100000ef9 <main+1>: mov rbp,rsp 0x0000000100000efc <main+4>: sub rsp,0x10 0x0000000100000f00 <main+8>: mov DWORD PTR [rbp-0x4],0xa 0x0000000100000f07 <main+15>: mov DWORD PTR [rbp-0x8],0x14 0x0000000100000f0e <main+22>: mov eax,DWORD PTR [rbp-0x4] 0x0000000100000f11 <main+25>: cmp eax,DWORD PTR [rbp-0x8] 0x0000000100000f14 <main+28>: jne 0x100000f22 <main+42> 0x0000000100000f16 <main+30>: lea rdi,[rip+0x19] # 0x100000f36 0x0000000100000f1d <main+37>: call 0x100000f30 <dyld_stub_puts> 0x0000000100000f22 <main+42>: mov eax,0x0 0x0000000100000f27 <main+47>: leave 0x0000000100000f28 <main+48>: ret We see that in the very basic version of this code with absolutely no optimizations and doing the silliest things that we can, we store our two values into some memory locations, perform a comparison (cmp), and jump if not equal. We can see that the jump leads us to the puts() call. Now, let's get smarter. The variables foo and bar do not change value, and we only work with two variables in the routine. Therefore, we could optimize by storing those values in temporary registers instead of using expensive memory transfers. Further,since our two constants are being compared and will always return a false, we actually have a section of code -- the printf -- that is dead code, that can be completely removed from final compilation. Well, that's simple, and everyone who uses C in production at least turns on some minor optimization: $ gcc -g -O1 example.c # the only difference is the -O1 Dump of assembler code for function main: 0x0000000100000f34 <main+0>: push rbp 0x0000000100000f35 <main+1>: mov rbp,rsp 0x0000000100000f38 <main+4>: mov eax,0x0 0x0000000100000f3d <main+9>: leave 0x0000000100000f3e <main+10>: ret This does not look like our C code at all! And thankfully so! What a waste of space and CPU time it would have been had we treated C like an interpreted language! C is a high-level language with numerous compiler implementations that can intelligently convert the human-readable code into the binary code that represents the real situation behind the code. The point here is that you are not properly guessing the assembler code that will be produced. The compiler is doing a better job of that; that is the compiler's job. As a programmer, you can just focus on the algorithm. C is not an assembler macro language. For that, you would use things like "gas".
- kevinnk 15y ago
  C is not assembly and hasn't been for a very long time. But I think when people use the people use the phrase "portable assembler" they really mean that in C you both control the memory layout of data types very finely and that code maps very directly to an equivalent assembly construct. True, optimizers frequently change the actual executed code from what what we expect, but C gives a very intuitive feel of what the "upper bound" assembly output is. For example in C "array[0] = (x + y);" will never be more than a couple assembly instructions long. In many languages, including Haskell (and in the case of operator overloading, C++), the equivalent construct might map to hundreds if not thousands of instructions. Or it might map to the same one or two that C would emit. It's impossible to know and there is no reasonable upper bound on what could happen.
  - stcredzero 15y ago
    might map to hundreds if not thousands of instructions. Or it might map to the same one or two that C would emit. It's impossible to know and there is no reasonable upper bound on what could happen. Over every possible piece of code that could be compiled anywhere, this might well be true. But for a properly informed programmer for a given piece of code, not so much.
    - kevinnk 15y ago
      >But for a properly informed programmer for a given piece of code, not so much. There are a couple reasons that even for "informed" programmers this is still important 1) For most dynamic languages, even simple operations can take a highly variable amount of time to execute. How many instructions does an array access take in Javascript? The answer depends on everything from the state of the JIT to the types involved, both of which are usually impossible to know before hand. In C we can answer this pretty easily. 2) The modern trend is towards writing more and more generic code. Even for statically compiled languages like C++ and Haskell, the actual underlying operations are purposely* abstracted away from you. Unless you know every possible instance that your code could be used it is impossible to know how long any operation will take. And all this is assuming that the programmer knows everything about their compiler, assembler, standard library, imported libraries, ect, which isn't true for all but the most expert programmers. *Admittedly, the actual length of time it takes is dependent on the state of the processor which can be very difficult to predict, but we will have a lot more information than we would have had otherwise.
Danieru 15y ago
"I know that the computer is comparing one value to another" Actually it isn't.

Xurinos 15y ago

This just isn't true. C is not portable assembler. It was never intended to be. I hear it claimed, and it is wrong every time somebody calls C a low-level language close to assembler. You can make some roughly reasonable assumptions about what comes out of the compiler, but often it is not what you think it is. Let's challenge this specific claim, that when you do "if (foo == bar)" -- I corrected the syntax error, which is a symptom of C's high-level syntax and not of the underlying assembly code -- you compare one value to another and then jump. For this challenge,I will write some trivial code that we should be able to make easy assumptions about, and I will compile it with debugging enabled so that I can dump the results with gdb. $ gcc -g example.c 1 #include <stdio.h> 2 3 int main() { 4 int foo = 10; 5 int bar = 20; 6 if (foo == bar) { 7 printf("Fun\n"); 8 } 9 return 0; 10 } Dump of assembler code for function main: 0x0000000100000ef8 <main+0>: push rbp 0x0000000100000ef9 <main+1>: mov rbp,rsp 0x0000000100000efc <main+4>: sub rsp,0x10 0x0000000100000f00 <main+8>: mov DWORD PTR [rbp-0x4],0xa 0x0000000100000f07 <main+15>: mov DWORD PTR [rbp-0x8],0x14 0x0000000100000f0e <main+22>: mov eax,DWORD PTR [rbp-0x4] 0x0000000100000f11 <main+25>: cmp eax,DWORD PTR [rbp-0x8] 0x0000000100000f14 <main+28>: jne 0x100000f22 <main+42> 0x0000000100000f16 <main+30>: lea rdi,[rip+0x19] # 0x100000f36 0x0000000100000f1d <main+37>: call 0x100000f30 <dyld_stub_puts> 0x0000000100000f22 <main+42>: mov eax,0x0 0x0000000100000f27 <main+47>: leave 0x0000000100000f28 <main+48>: ret We see that in the very basic version of this code with absolutely no optimizations and doing the silliest things that we can, we store our two values into some memory locations, perform a comparison (cmp), and jump if not equal. We can see that the jump leads us to the puts() call. Now, let's get smarter. The variables foo and bar do not change value, and we only work with two variables in the routine. Therefore, we could optimize by storing those values in temporary registers instead of using expensive memory transfers. Further,since our two constants are being compared and will always return a false, we actually have a section of code -- the printf -- that is dead code, that can be completely removed from final compilation. Well, that's simple, and everyone who uses C in production at least turns on some minor optimization: $ gcc -g -O1 example.c # the only difference is the -O1 Dump of assembler code for function main: 0x0000000100000f34 <main+0>: push rbp 0x0000000100000f35 <main+1>: mov rbp,rsp 0x0000000100000f38 <main+4>: mov eax,0x0 0x0000000100000f3d <main+9>: leave 0x0000000100000f3e <main+10>: ret This does not look like our C code at all! And thankfully so! What a waste of space and CPU time it would have been had we treated C like an interpreted language! C is a high-level language with numerous compiler implementations that can intelligently convert the human-readable code into the binary code that represents the real situation behind the code. The point here is that you are not properly guessing the assembler code that will be produced. The compiler is doing a better job of that; that is the compiler's job. As a programmer, you can just focus on the algorithm. C is not an assembler macro language. For that, you would use things like "gas".

kevinnk 15y ago
C is not assembly and hasn't been for a very long time. But I think when people use the people use the phrase "portable assembler" they really mean that in C you both control the memory layout of data types very finely and that code maps very directly to an equivalent assembly construct. True, optimizers frequently change the actual executed code from what what we expect, but C gives a very intuitive feel of what the "upper bound" assembly output is. For example in C "array[0] = (x + y);" will never be more than a couple assembly instructions long. In many languages, including Haskell (and in the case of operator overloading, C++), the equivalent construct might map to hundreds if not thousands of instructions. Or it might map to the same one or two that C would emit. It's impossible to know and there is no reasonable upper bound on what could happen.
- stcredzero 15y ago
  might map to hundreds if not thousands of instructions. Or it might map to the same one or two that C would emit. It's impossible to know and there is no reasonable upper bound on what could happen. Over every possible piece of code that could be compiled anywhere, this might well be true. But for a properly informed programmer for a given piece of code, not so much.
  - kevinnk 15y ago
    >But for a properly informed programmer for a given piece of code, not so much. There are a couple reasons that even for "informed" programmers this is still important 1) For most dynamic languages, even simple operations can take a highly variable amount of time to execute. How many instructions does an array access take in Javascript? The answer depends on everything from the state of the JIT to the types involved, both of which are usually impossible to know before hand. In C we can answer this pretty easily. 2) The modern trend is towards writing more and more generic code. Even for statically compiled languages like C++ and Haskell, the actual underlying operations are purposely* abstracted away from you. Unless you know every possible instance that your code could be used it is impossible to know how long any operation will take. And all this is assuming that the programmer knows everything about their compiler, assembler, standard library, imported libraries, ect, which isn't true for all but the most expert programmers. *Admittedly, the actual length of time it takes is dependent on the state of the processor which can be very difficult to predict, but we will have a lot more information than we would have had otherwise.

Danieru 15y ago

"I know that the computer is comparing one value to another" Actually it isn't.