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can any body give me the idea to solve this problem, not the whole answer.thanks, Use a matrix represents an island surrounded by water. Two bridges lead out of the island. A mouse is placed on the black square. Write a program to make the mouse take a walk across the island. The mouse is...

this is the output expression: - + * 9 + 2 8 * + 4 8 6 3 expression: + * 9 + 2 8 * + 4 8 6 3 - expression: * 9 + 2 8 * + 4 8 6 3 - + expression: 9 + 2 8 * + 4 8 6 3 - + * expression: + 2 8 * + 4 8 6 3 - + * 9 after calculating 2 + 8 = 10 expression: * + 4 8 6 3 - + * 9 10...

sorry guys for my stupid questions. i got the answer . code is #include <stdio.h> #include <stdlib.h> #include<ctype.h> #include <string.h> typedef struct node { char data[16]; struct node *link; } NODE; typedef struct queue { NODE *front; NODE *rear; int count; } QUEUE...

#include <stdio.h> #include <stdlib.h> #include<ctype.h> #include<math.h> typedef struct node { char data[8]; struct node *link; } NODE; typedef struct queue { NODE *front; NODE *rear; int count; } QUEUE; QUEUE* CreateQueue() { QUEUE* q =...

is converting from a char array to a single char possible ?? char data[]="163" to char data=163 and how to convert from int a=163 to char data[]="163" and int n =163 to char a = 163 thanks

actually i am trying to calculate the expression like this 1 .first take the char in expr[]="-+*9+28*+4863 ", calculate and put it in a queue . 2. again put t items in the queue to the expr[]= "-+*910*1263". 3. and repeat 1 and 2 till q->count is 1. now i have a doubt that my approach is...

i dint really get what you are saying . could u please explain me how to use it ??

so here is one more question , how to convert "expr[j]" and "expr[k]" to ctype strings and pass to "calculate" function ??

this is the question: One way to evaluate a prefix expression is to use a queue. To evaluate the expression, scan it repeatedly until the final expression value is known. In each scan, read the tokens and store them in a queue. In each scan, replace an operator followed by two operands by the...

is there any possibility to convert a char to a char array example : here is char a ='163'; char data[16]; i want to input the value of 'a' to 'data' as data[0]='1' data[1]='6' data[2]='3' data[3]='\0'

i want to pass two characters("6","8") into a function which will convert this two chars into integers and add these two integers(6+8=14) after that it will change the value (int 14) into char (char "14") and returns the char. can anybody help me please ??

Thank you ArkM, I got the solution as follows ________________________________ #include <stdio.h> #include<stdlib.h> #include<string.h> void newStrCpy(char* q, char* p); int main (void) { char *src; printf("\n----------------------------------------"); printf("\nPROGRAM TO COPY...

then how its working ??? int main() { char *p = "CUCKOO"; printf("%s\n",p); return 0; } output: CUCKOOhttp://www.mediafire.com/i/?gcgs17jhnn777h1

Thanks xwb , i tried by char *temp = (char*) malloc (strlen(q) + 1); but still the output is "null".. in my code i checked the value of "p" before going out of the function by debugging. at this point "p" is pointing to "temp". But once the compiler goes out of the function p is pointing...

what is wrong with this code? its not working . can anybody explain in detail please ??? ____________________________ #include <stdio.h> #include<stdlib.h> #include<string.h> void newStrCpy(char* q, char* p); int main (void) { char *q = "MANICHANDRA"; char *p ...

output of this _____________________ int main() { char s[]="MANI"; char *p; p=s; *p = "CUCKOO"; //*(p+1)='C'; //*(p+2) = 'B'; //*(p+3) = 'D'; printf("%s\n",s); printf("%s\n",p); return 0; } ____________________________ is eANI eANI y can't...

why can't we do this ?? int main() { char *p="MANI";; *(p+1)='C'; *(p+2) = 'B'; *(p+3) = 'D'; printf("%s\n",p); return 0; } when we can do this .. int main() { char s[]="MANI"; char *p; p=s; *(p+1)='C'; *(p+2) = 'B'; *(p+3) = 'D'...

output: M A N I 77 65 78 73 77 65 78 78 79 73 80 M A N N O I P MANI ANI NI I MANI ANI NI I 3844 3845 3846 3847 3844 3845 3846 3847 3844 MANI 3845 ANI

int main() { char *p="MANI"; char *q; printf("%c %c %c %c\n",'M','A','N','I'); printf("%d %d %d %d\n",'M','A','N','I'); printf("%d %d %d %d %d %d %d\n", *p,*(p+1),*p+1,*(p+2),*p+2,*(p+3),*p+3); printf("%c %c %c %c %c %c %c\n"...