Concatenation: Difference between revisions

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[[String]] concatenation is the operation of joining two or more character strings end-to-end. For example, the strings "snow" and "ball" may be concatenated to give "snowball".
[[String]] '''concatenation''' is the operation of joining two or more character strings end-to-end. For example, the strings "snow" and "ball" may be concatenated to give "snowball".


In BR, the operator responsible for string concatenation is the ampersand '''&'''. So, the "snowball" example above may be carried out as follows:
In BR, the operator responsible for string concatenation is the ampersand '''&'''. So, the "snowball" example above may be carried out as follows:
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===Appending===
===Appending===
{{Append}}
{{:Append}}


===Prepending===
===Prepending===
{{Prepend}}
{{:Prepend}}


<noinclude>
<noinclude>

Latest revision as of 18:37, 13 January 2014

String concatenation is the operation of joining two or more character strings end-to-end. For example, the strings "snow" and "ball" may be concatenated to give "snowball".

In BR, the operator responsible for string concatenation is the ampersand &. So, the "snowball" example above may be carried out as follows:

00010 let string1$ = "snow"
00020 let string2$ = "ball"
00030 let result$ = string1$ & string2$
00040 print result$ ! this will print snowball

You may concatenate as many strings as you like. Note that if you are storing the concatenated result into a string variable, then this variable needs to be dimensioned long enough to fit the combined length of all the concatenated strings. Consider the following example:

00010 let string1$ = "snow"
00020 let string2$ = "ball"
00030 let string3$ = " effect of concatenating many strings"
00040 let result$ = string1$ & string2$ & string3$ ! this will result in error

The above example results in an error, because the default length of the string result$ is 18 characters, as any BR string. The combined length of string1$, string2$, and string3$ is 45 characters. Note that the error does not result from concatenating string1$ & string2$ & string3$. The error occurs when we try to assign a 45 character value to an 18 character string result$. In order to correct the error, we need to dimension result$ to at least 45 characters or more. Below is the corrected example:

00005 dim result$*45
00010 let string1$ = "snow"
00020 let string2$ = "ball"
00030 let string3$ = " effect of concatenating many strings"
00040 let result$ = string1$ & string2$ & string3$ ! this will result in error

Advanced

For increased speed of program execution, and shorter syntax you may incorporate the following methods.

Appending

To append string2$ to string1$ means to join string2$ to the end of string1$.

To append to the end of a String you should (for maximum speed of code execution) use

X$(inf:0)="append this to end"

OR

X$(inf:inf)="append this to the end"

Here, inf denotes infinity.

So X$(inf:inf) means "the substring of X$ starting at infinity". This is particularly useful when you don't know how long your string is and do not want to calculate its length.

see also: prepend


Prepending

To prepend string1$ to string2$ means to join string1$ to the beginning of string2$.

To append to the beginning of a string you should use

X$(0:0)="append this to front"

or alternately

X$(1:0)="append this to front"

For example,

00010 dim result$*255 ! dimension long enough to fit the result
00020 let result$ = " and this is the end"
00030 let string_to_prepend$ = "this is the front"
00040 let result$(0:0) = string_to_prepend$
00050 print result$

The output of the program above will be:

this is the front and this is the end