我不确定使用浮点数是否是解决问题的***方法.但是听起来您尝试了所有其他选项，但非线性正逐渐成为您的障碍.请注意，即使使用浮点数对问题进行建模，浮点算法也非常棘手，求解器可能很难找到令人满意的模型.此外，由于数值不稳定，解决方案可能与实际结果相去甚远.

I'm not sure if using floats is the best way to go for your problem. But sounds like you tried all other options and non-linearity is getting in your way. Note that even if you model your problem with floats, floating-point arithmetic is quite tricky and solver may have hard time finding satisfying models. Furthermore, solutions maybe way far off from actual results due to numerical instability.

撇开所有这些，使用C api编写查询代码的正确方法是(假设我们使用32位单精度浮点数):

Leaving all those aside, the correct way to code your query using the C api would be (assuming we use 32-bit single-precision floats):

#include <z3.h>

int main(void) {
    Z3_config  cfg = Z3_mk_config();
    Z3_context ctx = Z3_mk_context(cfg);
    Z3_solver  s   = Z3_mk_solver(ctx);

    Z3_solver_inc_ref(ctx, s);
    Z3_del_config(cfg);

    Z3_sort float_sort = Z3_mk_fpa_sort(ctx, 8, 24);

    Z3_symbol s_x = Z3_mk_string_symbol(ctx, "x");
    Z3_symbol s_y = Z3_mk_string_symbol(ctx, "y");
    Z3_ast      x = Z3_mk_const(ctx, s_x, float_sort);
    Z3_ast      y = Z3_mk_const(ctx, s_y, float_sort);

    Z3_symbol s_x_times_y = Z3_mk_string_symbol(ctx, "x_times_y");
    Z3_ast      x_times_y = Z3_mk_const(ctx, s_x_times_y, float_sort);

    Z3_ast c1 = Z3_mk_eq(ctx, x_times_y, Z3_mk_fpa_mul(ctx, Z3_mk_fpa_rne(ctx), x, y));

    Z3_ast c2 = Z3_mk_fpa_gt(ctx, x_times_y, Z3_mk_fpa_numeral_float(ctx, 10, float_sort));

    Z3_solver_assert(ctx, s, c1);
    Z3_solver_assert(ctx, s, c2);

    Z3_lbool result = Z3_solver_check(ctx, s);

    switch(result) {
        case Z3_L_FALSE: printf("unsat\n");
                         break;
        case Z3_L_UNDEF: printf("undef\n");
                         break;
        case Z3_L_TRUE:  { Z3_model m = Z3_solver_get_model(ctx, s);
                           if(m) Z3_model_inc_ref(ctx, m);
                           printf("sat\n%s\n", Z3_model_to_string(ctx, m));
                           break;
                         }
    }

    return 0;
}

运行时，将打印:

sat
x_times_y -> (fp #b0 #xbe #b10110110110101010000010)
y -> (fp #b0 #xb5 #b00000000000000000000000)
x -> (fp #b0 #x88 #b10110110110101010000010)

这些是单精度浮点数；例如，您可以在Wikipedia中阅读有关它们的信息.在更常规的表示法中，它们是:

These are single-precision floating point numbers; you can read about them in wikipedia for instance. In more conventional notation, they are:

x_times_y -> 1.5810592e19
y         -> 1.8014399e16
x         -> 877.6642

使用起来很棘手，但是您要问什么.

This is quite tricky to use, but what you have asked.

在向此类复杂的C代码投资之前，我衷心建议使用Python API，至少要了解求解器的功能.这是在Python中的外观:

I'd heartily recommend using the Python API to at least see what the solver is capable of before investing into such complicated C code. Here's how it would look in Python:

from z3 import *

x = FP('x', FPSort(8, 24))
y = FP('y', FPSort(8, 24))

s = Solver()
s.add(x*y > 10);
s.check()
print s.model()

运行时，将打印:

[y = 1.32167303562164306640625,
 x = 1.513233661651611328125*(2**121)]

也许不是您期望的那样，但这确实是一个有效的模型.

Perhaps not what you expected, but it is a valid model indeed.

仅是为了给您带来简单的感觉，这是使用Haskell绑定可以表达相同问题的方式(这仅仅是一个衬里！)

Just to give you a taste of simplicity, here's how the same problem can be expressed using the Haskell bindings (It's just a mere one liner!)

Prelude Data.SBV> sat $ \x y -> fpIsPoint x &&& fpIsPoint y &&& x * y .> (10::SFloat)
Satisfiable. Model:
  s0 = 5.1129496e28 :: Float
  s1 =  6.6554557e9 :: Float

摘要

请注意，浮点还存在与NaN/Infinity值有关的问题，因此您可能必须明确避免使用这些值. (这是Haskell表达式通过使用isFPPoint谓词所做的操作.用Python或C对其进行编码将需要更多代码，但肯定是可行的.)

Summary

Note that Floating-point also has issues regarding NaN/Infinity values, so you might have to avoid those explicitly. (This is what the Haskell expression did by using the isFPPoint predicate. Coding it in Python or C would require more code, but is surely doable.)

应该强调的是，实际上，与Z3的任何其他绑定(Python，Haskell，Scala，还有您所拥有的东西)都将为您提供比C/C ++/Java更好的体验. (即使在SMTLib中直接编码也会更好.)

It should be emphasized that literally any other binding to Z3 (Python, Haskell, Scala, what have you) will give you a better experience than what you'll get with C/C++/Java. (Even direct coding in SMTLib would be nicer.)

因此，我衷心建议使用一些更高级别的接口(Python是一个很好的接口:它很容易学习)，一旦您对模型及其工作方式充满信心，就可以开始在C语言中对其进行编码如有必要.

So, I heartily recommend using some higher-level interface (Python is a good one: It is easy to learn), and once you are confident with the model and how it works, you can then start coding the same in C if necessary.