error output on incrementing

[crazysomesh007]

hai,
iam a new to C and also new to this site , please give me the explaination to the following program why the out put is 56.

int i=7;
printf("%d",i++*i++);

[horace1]

hai,
iam a new to C and also new to this site , please give me the explaination to the following program why the out put is 56.

int i=7;
printf("%d",i++*i++);

with i++ the value of the operand is used in the expression then it is incremented. e.g. assuming i = 10

   k = i++; after assignment k would = 10 and i = 11
 

Remember, in addition to evaluating the expression, the value of the variable in memory is altered (this is called a side effect). Do not use a variable more than once in an expression if one (or more) of the references has one of these operators attached to it. The C/C++ standard does not specify the order in which the operands of an operator are evaluated and there is no guarantee when an affected variable will change its value.

so in your example

i++*i++
 

when is the ++ done? you don't know. Different compilers give different results.

also see post
http://www.thescripts.com/forum/thread570241.html

[smithan]

hai,
iam a new to C and also new to this site , please give me the explaination to the following program why the out put is 56.

int i=7;
printf("%d",i++*i++);

hi,

u must have guessed that the output is 56 because it is taking the value of 7*8 ..... now when "i" has the value 7 it multiplies with 8 because, of the expression i++......

i++, "++" in this expression increaments the current value of i by "1" , but it does so only after "i" is used......

so, this is what is happening i ur program....

i=7; // the value of "i" is set to "7" .....
printf("%d",i++*i++); // i is 7 first n then it is increamented hence it becomes 8, so, 7*8 is taken as during the second occurance of "i" its value would already be incremented........ this is how the statements look after execution....

i=7
printf("%d", 7*8);
// i =9 at the end, coz u have increamented i in its last occurance.....

hence the result being 56

smitha

[smithan]

hai,
iam a new to C and also new to this site , please give me the explaination to the following program why the out put is 56.

int i=7;
printf("%d",i++*i++);

and also if ur trying to print the value of 7* 7 then u have to use
printf("%d",i*i);

and incase u want to increment the value of i after multiplying then use
printf("%d",i*i++);

smitha

[horace1]

and also if ur trying to print the value of 7* 7 then u have to use
printf("%d",i*i);

and incase u want to increment the value of i after multiplying then use
printf("%d",i*i++);

smitha

Not advisable! remember in addition to evaluating the expression i*i the value of the variable i in memory is altered (this is called a side effect). The standard does not specify the order in which the operands of an operator are evaluated and there is no guarantee when an affected variable will change its value. For example

    int i=7;
    printf("%d",i*i++);
 

prints
49
using DEV-C++ with the GNU gcc compiler and
56
when using Turbo C V3.01

simillarly your original code

int i=7;
printf("%d",i++*i++);
 

prints
49
using DEV-C++ with the GNU gcc compiler and
56
when using Turbo C V3.01

Do not use a variable more than once in an expression if one (or more) of the references has i++ or i-- attached to it. You don't know when the side effects takes place.