在 C 语言中计算以下表达式时，输出为零而不是无穷大。但根据 C 运算符优先级规则，输出应该是无穷大。
double a=1/(1.0/0.0);
printf("a : %.18le\n", a);
请解释一下 gcc 编译器如何评估它？
最佳答案
C 标准没有规定 double 如何处理 NaN 和 Inf 数字，但是对于 gcc，其行为由严格模式下的 IEEE 754 规定:https://en.wikipedia.org/wiki/IEEE_754
摘自那篇文章:
The standard defines five exceptions, each of which returns a default value and has a corresponding status flag that (except in certain cases of underflow) is raised when the exception occurs. No other exception handling is required, but additional non-default alternatives are recommended (see below).
The five possible exceptions are:
Invalid operation: mathematically undefined, e.g., the square root of a negative number. Returns qNaN by default.
Division by zero: an operation on finite operands gives an exact infinite result, e.g., 1/0 or log(0). Returns ±infinity by default.
- Overflow: a result is too large to be represented correctly (i.e., its exponent with an unbounded exponent range would be larger than emax). Returns ±infinity by default for the round-to-nearest mode.
- Underflow: a result is very small (outside the normal range) and is inexact. Returns a subnormal or zero by default.
- Inexact: the exact (i.e., unrounded) result is not representable exactly. Returns the correctly rounded result by default.
但是，在某些平台上，IEEE 754 兼容的浮点单元不可用，您应该强制使用软件浮点库或查阅平台手册以了解发生的情况。例如，arm 的 fpu 有一个“RunFast”模式，该模式会禁用严格合规性。
关于c - 1/(1.0/0.0) 在 C 中计算为零，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/50176461/