C (programming language)/Tutorials: Difference between revisions
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imported>Eric Evers (New page: =C Programming Language Tutorials= ==Truth values and Iverson Bracket== The C programming language is set up nicely to support the Iverson Bracket set notation. Iverson proposed that tr...) |
imported>Eric Evers No edit summary |
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==Truth values and Iverson Bracket== | ==Truth values in C and the Iverson Bracket== | ||
The C programming language is set up nicely to support the Iverson Bracket set notation. Iverson proposed that truth values | Truth values in C can be simple integers. The C programming language is set up nicely to support the Iverson Bracket set notation. Iverson proposed that truth values should be numerical with true=1 and false=0. Actually 0=false in C and all non-zero values are True, even -1 is true in C. | ||
zero : false | |||
non-zero : true | |||
The following calculates the maximum of two numbers using an Iverson bracket formula | The following calculates the maximum of two numbers using an Iverson bracket formula in C. | ||
The traditional Iverson bracket notation in algebra: [a>b] gives 1 or 0. | |||
max = (a>=b)*a + (b<a)*b; | max = (a>=b)*a + (b<a)*b; in C | ||
============================= lets | ============================= lets calculate using 6 and 7 | ||
7 = (6>=7)*6 + (7<6)*7 | 7 = (6>=7)*6 + (7<6)*7 in algegra |
Revision as of 20:30, 22 April 2008
C Programming Language Tutorials
Truth values in C and the Iverson Bracket
Truth values in C can be simple integers. The C programming language is set up nicely to support the Iverson Bracket set notation. Iverson proposed that truth values should be numerical with true=1 and false=0. Actually 0=false in C and all non-zero values are True, even -1 is true in C.
zero : false non-zero : true
The following calculates the maximum of two numbers using an Iverson bracket formula in C.
The traditional Iverson bracket notation in algebra: [a>b] gives 1 or 0.
max = (a>=b)*a + (b<a)*b; in C ============================= lets calculate using 6 and 7 7 = (6>=7)*6 + (7<6)*7 in algegra