I have three Pointers:
int *x,*y, *Temp; And then I have done
x = (int *)malloc(sizeof(int)*m); y = (int *)malloc(sizeof(int)*n); Temp = (int *)malloc(sizeof(int)*(m+n)); where m and n are certain values.
Next, I have entered values into Temp.
for(i=0; i < m+n; i++) { scanf("%d", Temp+i); } I want half of Temp in x and the other half in y. How do I do this?
for(i=0; i < m; i++) { x[i] = Temp[i]; } The code above to copy the contents is not working!
Also, how do I print the values?
Using memcpy is probably the easiest way to accomplish what you want:
memcpy(x, Temp, m * sizeof*Temp); memcpy(y, &Temp[m], n * sizeof*Temp); To print the values, just use printf:
puts("Values in x:"); for(int i = 0; i < m; ++i) { printf("%d\n", x[i]); } puts("Values in y:"); for(int i = 0; i < n; ++i) { printf("%d\n", y[i]); } 
memcpy and printf in a loop". I moved printf to a different section, but neglected to remove the "in a loop" part. I fixed the answer.You can use as follows to print the values. I just use M in place of m for better clarity.
for(i=0; i < M; i++) { printf("VAL: %d\n",x[i]); } Similarly you can also print another array y.
Use memcpy.It will copy desired numbre of bytes .
What I was doing is already mentioned in this answer of PC Luddite
It doesn't really look like you need to allocate all that memory, nor does it seem like you need to copy it. Consider using this alternative approach:
int* xy = malloc(sizeof(*xy) * (m+n)); int* x = xy; int* y = xy + n; ... free(xy); 
malloc(or any function returningvoid *).2to the size ofTemp?callocwhich initializes the allocated memory or usememsetto set the memory locations to NULL.