Linux上printf的分段错误

Question

@Team,

While trying to print an integer value through printf i accidentally wrote the statement as

int x =10;
printf(x);

In linux i am getting a segmentation fault when tried to execute it. Although its wrong but can some one please help me to know the reason for it.

Strace says:

mprotect(0x7f872fb26000, 4096, PROT_READ) = 0
munmap(0x7f872fb0b000, 99154)           = 0
--- SIGSEGV (Segmentation fault) @ 0 (0) ---
+++ killed by SIGSEGV (core dumped) +++

Tried searching in SO but to no success.

Answer 1

First parameter for printf() is format string which is char * pointer. So when you do printf(x) it takes x as char * and tries to access string stored at address 10 . But its invalid so it gives segmentation fault.

Answer 2

之所以会出现分段错误，是因为printf会将通过它的10解释为char * ，并试图从计算机地址10读取。在运行Linux的系统上，该地址无效并导致分段错误。

Answer 3

Use printf() in proper format. printf("%d",x).