How does one remove all whitespaces in an alphanumeric COBOL string?

Answer 1

Frankly, you can't, at least not with a single statement. Since COBOL automatically pads the rightmost characters in an alphanumeric field, an INSPECT might left-justify and pad the characters with a statement like this:

INSPECT some-field REPLACING ALL SPACES BY ''. (assumes by 'whitespace' you only mean spaces. If not, you would need additional REPLACING clauses for other characters to bypass.)

However, a way guaranteed to work is to copy non-whitespace characters into another field.

77 whitespace-test PIC X.

88 is-whitespace VALUES ' ', ... (whatever additional characters you want to remove)

77 char-count PIC S9(4) COMP.

77 sub1 PIC S9(4) COMP.

01 work-fields.

05 source-field PIC X(something).

10 source-field-char REDEFINES source-field PIC X OCCURS something TIMES.

05 target-field PIC X(something) VALUE SPACES.

10 target-field-char REDEFINES target-field PIC X OCCURS something TIMES.

MOVE ZERO TO char-count.

PERFORM VARYING sub1 FROM 1 BY 1 UNTIL sub1 EQUAL something

MOVE source-field-char(sub1) TO whitespace-test

IF NOT is-whitespace

ADD 1 TO char-count

MOVE whitespace-test TO target-field-char(char-count)

END-IF

END-PERFORM.