I am getting the below error, what is std=c99/std=gnu99 mode?
source Code:
#include <stdio.h> void funct(int[5]); int main() { int Arr[5]={1,2,3,4,5}; funct(Arr); for(int j=0;j<5;j++) printf("%d",Arr[j]); } void funct(int p[5]) { int i,j; for(i=6,j=0;i<11;i++,j++) p[j]=i; } Error Message: hello.c: In function ‘main’: hello.c:11:2: error: ‘for’ loop initial declarations are only allowed in C99 mode for(int j=0;j<5;j++) ^ hello.c:11:2: note: use option -std=c99 or -std=gnu99 to compile your code` 4 Answers
This happens because declaring variables inside a for loop wasn't valid C until C99(which is the standard of C published in 1999), you can either declare your counter outside the for as pointed out by others or use the -std=c99 flag to tell the compiler explicitly that you're using this standard and it should interpret it as such.
You need to declare the variable j used for the first for loop before the loop.
int j; for(j=0;j<5;j++) printf("%d",Arr[j]); Easiest Solution by "Prof. Dr. Michael Helbig" . it will switch your mode to c99 so you don't have to add flag every time in make file
Solution: use the option -std=c99 for your compiler! Go to: Project > Properties > C/C++ Buils > Settings > Tool Settings > GCC C Compiler > Dialect > Language Standard: choose "ISO C99"
This will be working code
#include <stdio.h> void funct(int[5]); int main() { int Arr[5]={1,2,3,4,5}; int j = 0; funct(Arr); for(j=0;j<5;j++) printf("%d",Arr[j]); } void funct(int p[5]){ int i,j; for(i=6,j=0;i<11;i++,j++) p[j]=i; } 
