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TEST1-5
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1985-12-30
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436 lines
.T
and now a TEST
.R4C1
1 ~b~Imain() { ~N
2 ~b~I char key; ~N
3 ~b~I printf("\n Press a key "); ~N
4 ~b~I scanf("%c",key); ~N
5 ~b~I printf("\n You pressed %c",key);~N
6 ~b~I} ~N
.Q
Which line has the error?
4
Line ~I4~N should read:
4 ~b~I scanf("%c",~F&~N~b~Ikey); ~N
.K19,60
I
know
that!
.WN
1 ~b~Imain() { ~N
2 ~b~I char key; ~N
3 ~b~I printf("\n Press a key "); ~N
4 ~b~I scanf("%c",&key); ~N
5 ~b~I printf("\n You pressed %c",key) ~N
6 ~b~I} ~N
.Q
Which line has the error?
5
Line ~I5~N should read:
5 ~b~I printf("\n You pressed %c",key)~F;~N
.K19,60
;;;;;;;;;;
.WN
.Q
If x is an integer variable, then where in memory is it located?
&x
The ~Iaddress~N, in memory, of any variable ~b~Ix~N is ~b~I&x~N
.K16,30
&x=
address
.R9C1
.Q
&x is called a ?
pointer
.K16,30
pointer
to
x
.WN
1 ~b~Imain() { ~N
2 ~b~I int i; ~N
3 ~b~I char x='A'; ~N
4 ~b~I for (i=0; i<26; i++,x++) ~N
5 ~b~I printf("%c",x); ~N
6 ~b~I} ~N
.Q
How many errors can you find ?
0
.R8C60
~INO ERRORS!~N
.R11C1
4 ~b~I for (i=0; i<26; i++,~Fx++~N~b~I) ~N
Although ~b~Ix~N is declared a ~b~Ichar~N and given the 'value' ~b~IA~N,
the actual number stored in memory is (probably!) ~I65~N (in decimal)
which is the ASCII value for an ~IA~N. This ~I65~N is incremented via ~b~Ix++~N
and ~b~Iprintf()~N will print: ~r~IABCDEFGHIJKLMNOPQRSTUVWXYZ~N
.K19,60
probably??
.W
.R7C1B
~V DON'T TRUST THE INTERNAL COMPUTER 'VALUE' TO BE ASCII!!
.W
.R7C1B
~V however, if x is the 'A'-value, x+1 SHOULD be the 'B'-value...right?
.WK8,30
RIGHT!
.WN
1 ~b~Imain() { ~N
2 ~b~I printf("%s,%s,%s", ~N
3 ~b~I "Now is the time", ~N
4 ~b~I "for all good me", ~N
5 ~b~I "n to come to th", ~N
6 ~b~I "e aid of thier ", ~N
7 ~b~I "country."; ~N
8 ~b~I} ~N
.Q
How many errors here?
2
~V i-before-e-except-after-c ? ~N
.R6C1
6 ~b~I "he aid of ~Fthier~N~b~I", ~N
.K2,50
their
.WR14C1
...and the printout would be:
~r~INow is the timefor all good men to come to the aid of thier country.~N
tch!tch!
.WN
1 ~b~Imain() { ~N
2 ~b~I int w; ~N
3 ~b~I char z='z'; ~N
4 ~b~I w=z; ~N
5 ~b~I while (z>'w'); ~N
6 ~b~I printf("%c=%d;",z--,w--);~N
7 ~b~I} ~N
.Q
Which line has the error ?
5
Line 5 shouldn't have a semi-colon! (NOT after ~Iwhile~N or ~Ifor~N):
.R5C1
5 ~b~I while (z>'w')~F;~N~b~I ~N
.WK2,50
z--,w--
???
.R14C1
Line 4 sets ~b~Iw~N to the ~Ivalue~N of ~b~Iz~N ( 122 in ASCII ).
Line 6 is executed, ~b~Iprintf~N-ing the ~b~I%c~Nhar ~b~Iz~N and the ~b~I%d~Necimal
~b~Iw~N, then both ~b~Iz~N and ~b~Iw~N are ~IDECREMENTED~N (via ~b~Iz--,w--~N),
and line 6 is repeated ...as long as ~b~I%c~Nharacter ~b~Iz~N is greater than
~b~I'w'~N (meaning the 'value' of ~b~Iz~N exceeds the 'value' of the character
~b~Iw~N, which, in ASCII, is 119).
.W
The printout would be:
~r~Iz=122;y=121;x=120;~N
.WK2,50
confusing!
.WN
1 ~b~Imain() { ~N
2 ~b~I char s; ~N
3 ~b~I s="I'm a string"; ~N
4 ~b~I printf("%s",&s); ~N
5 ~b~I} ~N
.Q
Which line has the error(s) ?
4
Line 4 should be:
4 ~b~I printf("%s",s); ~N ...without the ~b~I&~N !
.R13C1
Although ~b~Iprintf()~N expects to recieve the ~Iaddress~N of a ~b~I%s~Ntring
variable (such as ~b~Is~N), we must ~V let C do it~N! We just refer to ~b~Is~N,
in ~b~Iprintf("%s",s)~N, and the C-compiler will look after passing the address
to ~b~Iprintf()~N.
.W
.R13C1
Although ~b~Iprintf()~N expects to ~Frecieve~N the ~Iaddress~N of a ~b~I%s~Ntring
.K2,50
sorry!
.WR13C1
Although ~b~Iprintf()~N expects to receive the ~Iaddress~N of a ~b~I%s~Ntring
.R18C1
The printout might be:
~r~I∞I~N ...or some other garbage!
.WNT
single CHARacters and INTegers are the same ?
.R4C1
1 ~b~Imain() { ~N
2 ~b~I char x; ~N
3 ~b~I for (x=0; x<=255; x++) ~N
4 ~b~I printf("%c",x); ~N
5 ~b~I} ~N
.Q
Will this print the IBM PC character set ..once ?
n
When we increment ~b~Ix~N from ~b~I0~N to ~b~I255~N we ~IDO~N get a
printout of the ~b~I%c~Nharacters whose 'values' are ~I0~N, ~I1~N, ~I2~N, etc.
BUT, because a ~b~I%c~Nharacter ~IALWAYS~N occupies a single byte in memory
their 'values' are ~IALWAYS~N ~b~I<=255~N (less-than-or-equal-to 255).
SO ...incrementing ~b~Ix~N, when it has the 'value' ~I255~N, will give the 'value'
~b~I0~N ...and the above program will continue to cycle through the character
set ...forever!
.K4,65
forever?
.W
1 ~b~Imain() { ~N But ~Ithis~N program WILL stop
2 ~b~I int x; ~N when the ~b~I%int~Neger ~b~Ix~N
3 ~b~I for (x=0; x<=255; x++) ~N reaches the value ~I256~N, since
4 ~b~I printf("%c",x); ~N ~b~Iint~Negers CAN be larger than ~I255~N
5 ~b~I} ~N
.K4,65
int is big
.WN
1 ~b~Imain() { ~N
2 ~b~I float x, y; ~N
3 ~b~I printf("Enter two numbers); ~N
4 ~b~I scanf("%f%f",x,y); ~N
5 ~b~I printf("%s%f", ~N
6 ~b~I "Their sum is ", ~N
7 ~b~I sum(x,y); ~N
8 ~b~I} ~N
9 ~b~Isum(a,b) ~N
10~b~I{ ~N
11~b~Ifloat a, b; ~N
12~b~Ireturn(a+b); ~N
13~b~I} ~N
.Q
How many BUGS!@#$ ???
4
.WR18C1
Forgot something in line 3....
.R3C38
~b~I printf("Enter two numbers~F"~N~b~I); ~N
.WR18C1
Forgot something in line 7 too!
.R7C38
~b~I sum(x,y)~F)~N~b~I; ~N
.WR18C1
Forgot to declare the function type (else we'd get ~Iintegers~N returned!)
.R9C38
~b~I~Ffloat~N~b~I sum(a,b) ~N
.WR18C1
.R18C1
We ~IMUST~N declare function arguments ~IIMMEDIATELY~N (~Ibefore~N the ~b~I{~N)!
.R10C38
~b~Ifloat a, b; ~N wrong
.R11C38
~b~I{ ~N order
.WN
1 ~b~Imain() { ~N
2 ~b~I int words=0; ~N
3 ~b~I char c; ~N
4 ~b~I while ( (c=getchar())!='\r' && c!='\n') }~N
5 ~b~I if (c == ' ') ~N
6 ~b~I words++; ~N
7 ~b~I } ~N
8 ~b~I printf("\n Number = %d",words); ~N
9 ~b~I} ~N
.Q
Any errors here (y/n) ?
n
.R10C60
~INO ERRORS~N
.R4C1
4 ~b~I while ( (~Vc=getchar()~N~b~I)!='\r' && c!='\n') }~N
.R13C1
Here we invoke the C-library function ~b~Igetchar()~N which inputs a
~Isingle~N character, which we assign to our ~b~Ichar~Nacter variable ~b~Ic~N.
.WR4C1
4 ~b~I while ( (c=getchar())~V!='\r' && c!='\n'~N~b~I) }~N
.R13C1
.R13C1
Here we check to see that the ~b~Ic~Nharacter is NOT EQUAL (~b~I!=~N) to a
~b~I\r~Neturn AND (~b~I&&~N) NOT EQUAL to a ~b~I\n~Newline. ( This then
will exit the ~b~Iwhile~N loop when we press the ~IEnter~N key ).
As long as it's NOT EQUAL, we execute the ~b~Iwhile~N loop.
.K2,60
&&
=
AND
!!
.WR4C1
4 ~b~I while ( (c=getchar())!='\r' && c!='\n') }~N
5 ~V if (c == ' ') ~N
6 ~V words++; ~N
.R13C1
.R13C1
Inside the ~b~Iwhile~N, we check if the ~b~Ic~Nharacter ~b~I== ' '~N
(~IEQUAL~N a space) and, if it is, we increment the ~b~Iint~Neger ~b~Iwords~N.
.WR5C1
5 ~b~I if (c == ' ') ~N
6 ~b~I words++; ~N
7 ~V } ~N
.R13C1
.R13C1
The first time that the ~IEnter~N key ~b~Ic~Nharacter occurs, we exit the
~b~Iwhile~N loop ...and this ~b~I}~N defines the ~Iend-of-the-while~N.
.WR7C1
7 ~b~I } ~N
8 ~V printf("\n Number = %d",words); ~N
.R13C1
.R13C1
After leaving the ~b~Iwhile~N, we ~b~Iprintf()~N the value of ~b~Iwords~N, which
may (or may not!) agree with the number of 'words' typed.
(It WILL if you ~Ifollow~N every word with ~Ione~N space!)
.WR8C1
8 ~b~I printf("\n Number = %d",words); ~N
9 ~V} ~N
.R13C1
.R13C1
...and the end of ~b~Imain()~N ( = end of the program! )
.WR15C1
If we compiled/linked/ran this program, we'd get:
~VThis is a line which has 7 words ~N You type this, and press ~IEnter~N.
~r~I Number = 8~N The computer responds thus!
.WK2,60
SMART!
.WN
Suppose we have 3 ~b~Ichar~Nacter variables: ~b~Ix='A' ; y='a' ; z='B'~N
.Q
Is x!=y (y/n)?
y
.Q
Is x!=z (y/n)?
y
.Q
Is x!=x (y/n)?
n
.Q
Is x!=y && x!=z (y/n)?
y
.WR15C1
Since x ~IIS~N different from y, the first answer is ~Iy~N.
Since x ~IIS~N different from z, the second answer is ~Iy~N.
Since x ~IIS NOT~N different from x, the third answer is ~In~N.
Since x ~IIS~N different from y ~IAND~N x ~IIS~N different from z
the fourth answer is ~Iy~N.
.WN
0 ~b~I#include <stdio.h> ~N
1 ~b~Imain() { ~N
2 ~b~Ifloat x=34.56; ~N
3 ~b~Iprintf("%6.1f",x); ~N
4 ~b~Iprintf("%-6.1f",x); ~N
5 ~b~Iprintf("%06.1f",x); ~N
6 ~b~I} ~N
.Q
What will be printed in Line 3 ?
34.6
.Q
What will be printed in Line 4 ?
34.6
.Q
What will be printed in Line 5 ?
0034.6
.WN
.Q
A C-expression meaning: n=n+1
n++
.Q
Another C-expression meaning: n=n+1
++n
.Q
A C-expression meaning: n=n+5
n+=5
.Q
A C-expression meaning: n=n/7
n/=7
.Q
A C-expression meaning: n=n-2
n-=2
.Q
A C-expression meaning: n=n*9
n*=9
.WN
~b~I if ( (x IS GREATER THAN 5) AND (x IS NOT EQUAL TO 7) ) ~N
~b~I if ( x IS EQUAL TO 9 ) ~N
.Q
What C operator should replace IS GREATER THAN ?
>
.Q
What C operator should replace AND ?
&&
.Q
What C operator should replace IS NOT EQUAL TO
!=
.Q
What C operator should replace IS EQUAL TO
==
.WN
1 ~b~Imain() { ~N
2 ~b~Iint i; ~N
3 ~b~Ifloat x; ~N
4 ~b~Ii=5.2; ~N
5 ~b~Ix=i/2; ~N
6 ~b~Iprintf("%f",x); ~N
7 ~b~I} ~N
.Q
What will be printed ?
2.000000
In line 4, ~b~I5.2~N (a ~b~Ifloat~N) is converted to an ~b~Iint~N 'cause ~b~Ii~N
is declared to be an ~b~Iint~N in Line 2, so the number ~I5~N is assigned to ~b~Ii~N.
In Line 5, because both ~b~Ii~N and ~b~I2~N are ~b~Iint~Negers ( no decinal in ~b~I2~N,
remember?), the division gives ~I5/2~N without the decimal part, ~I2~N, which
is then converted to ~b~Ifloat~N ('cause ~b~Ix~N is a ~b~Ifloat~N) and assigned
to ~b~Ix~N ....and ~b~Iprintf()~N gives 6 decimal places (unless told otherwise!).
.WN
.T
That's all folks!
.K16,32
au revoir!
.q