exp, expf, expl
From cppreference.com
| Defined in header
<math.h>
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||
| float expf( float arg );
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(1) | (since C99) |
| double exp( double arg );
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(2) | |
| long double expl( long double arg );
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(3) | (since C99) |
| Defined in header
<tgmath.h>
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| #define exp( arg )
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(4) | (since C99) |
1-3) Computes the e (Euler's number,
2.7182818) raised to the given power arg.
4) Type-generic macro: If
arg has type long double, expl is called. Otherwise, if arg has integer type or the type double, exp is called. Otherwise, expf is called. If arg is complex or imaginary, then the macro invokes the corresponding complex function (cexpf, cexp, cexpl).
Contents |
[edit] Parameters
| arg | - | floating point value |
[edit] Return value
If no errors occur, the base-e exponential of arg (earg
) is returned.
If a range error due to overflow occurs, +HUGE_VAL, +HUGE_VALF, or +HUGE_VALL is returned.
If a range error occurs due to underflow, the correct result (after rounding) is returned.
[edit] Error handling
Errors are reported as specified in math_errhandling.
If the implementation supports IEEE floating-point arithmetic (IEC 60559),
- If the argument is ±0, 1 is returned
- If the argument is -∞, +0 is returned
- If the argument is +∞, +∞ is returned
- If the argument is NaN, NaN is returned
[edit] Notes
For IEEE-compatible type double, overflow is guaranteed if 709.8 < arg, and underflow is guaranteed if arg < -708.4.
[edit] Example
Run this code
#include <stdio.h> #include <math.h> #include <float.h> #include <errno.h> #include <fenv.h> #pragma STDC FENV_ACCESS ON int main(void) { printf("exp(1) = %f\n", exp(1)); printf("FV of $100, continuously compounded at 3%% for 1 year = %f\n", 100*exp(0.03)); // special values printf("exp(-0) = %f\n", exp(-0.0)); printf("exp(-Inf) = %f\n", exp(-INFINITY)); //error handling errno = 0; feclearexcept(FE_ALL_EXCEPT); printf("exp(710) = %f\n", exp(710)); if(errno == ERANGE) perror(" errno == ERANGE"); if(fetestexcept(FE_OVERFLOW)) puts(" FE_OVERFLOW raised"); }
Possible output:
exp(1) = 2.718282
FV of $100, continuously compounded at 3% for 1 year = 103.045453
exp(-0) = 1.000000
exp(-Inf) = 0.000000
exp(710) = inf
errno == ERANGE: Numerical result out of range
FE_OVERFLOW raised[edit] References
- C11 standard (ISO/IEC 9899:2011):
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- 7.12.6.1 The exp functions (p: 242)
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- 7.25 Type-generic math <tgmath.h> (p: 373-375)
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- F.10.3.1 The exp functions (p: 520)
- C99 standard (ISO/IEC 9899:1999):
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- 7.12.6.1 The exp functions (p: 223)
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- 7.22 Type-generic math <tgmath.h> (p: 335-337)
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- F.9.3.1 The exp functions (p: 458)
- C89/C90 standard (ISO/IEC 9899:1990):
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- 4.5.4.1 The exp function
[edit] See also
| (C99)(C99)(C99)
|
computes 2 raised to the given power (2x) (function) |
| (C99)(C99)(C99)
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computes e raised to the given power, minus one (ex-1) (function) |
| (C99)(C99)
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computes natural (base-e) logarithm (ln(x)) (function) |
| (C99)(C99)(C99)
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computes the complex base-e exponential (function) |
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C++ documentation for exp
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