Consider the following C program: int fun (int *i) { *i += 5; return 4; } void m

Question

Consider the following C program: int fun (int *i) {    *i += 5;    return 4; } void main ( ) {    int x = 3;    x = x + fun(& x); } What is the value of x after the assignment statement in main,assuming    a. operands are evaluated left toright.    b. operands are evaluated right toleft. Consider the following C program: int fun (int *i) {    *i += 5;    return 4; } void main ( ) {    int x = 3;    x = x + fun(& x); } What is the value of x after the assignment statement in main,assuming    a. operands are evaluated left toright.    b. operands are evaluated right toleft.

Explanation / Answer

please rate - thanks What is the value of x after the assignment statement in main,assuming
   a. operands are evaluated left toright.    7    b. operands are evaluated right toleft.    12
left to right
int fun (int *i) {    *i +=5;          we add 5to x making x 8
   return4;        we return 4 tomain
} void main ( ) {    int x = 3;    x = x + fun(&x);     x starts at 3   in fun it ischanged to 8 but the 3 is already being used the 4 fromfun is added (3+4) to it making it 7
}
right toleft           x starts at 3 in fun x gets changed to 8 and fun is returned as 4so we have 4 + 8 = 12


to paraphrase
left to right it's is 3 + 4 =7           original value of x is used
right to left it is 4 + 8 =12           x is used as it was changed by fun, since fun was alreadyexecuted



What is the value of x after the assignment statement in main,assuming
   a. operands are evaluated left toright.    7    b. operands are evaluated right toleft.    12
left to right
int fun (int *i) {    *i +=5;          we add 5to x making x 8
   return4;        we return 4 tomain
} void main ( ) {    int x = 3;    x = x + fun(&x);     x starts at 3   in fun it ischanged to 8 but the 3 is already being used the 4 fromfun is added (3+4) to it making it 7
}
right toleft           x starts at 3 in fun x gets changed to 8 and fun is returned as 4so we have 4 + 8 = 12


to paraphrase
left to right it's is 3 + 4 =7           original value of x is used
right to left it is 4 + 8 =12           x is used as it was changed by fun, since fun was alreadyexecuted



int fun (int *i) {    *i +=5;          we add 5to x making x 8
   return4;        we return 4 tomain
} void main ( ) {    int x = 3;    x = x + fun(&x);     x starts at 3   in fun it ischanged to 8 but the 3 is already being used the 4 fromfun is added (3+4) to it making it 7
}
right toleft           x starts at 3 in fun x gets changed to 8 and fun is returned as 4so we have 4 + 8 = 12


to paraphrase
left to right it's is 3 + 4 =7           original value of x is used
right to left it is 4 + 8 =12           x is used as it was changed by fun, since fun was alreadyexecuted

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