c - 使用C99标准导致GCC编译错误

Question

我有一个名为“print.c”的非常简单的打印程序:

#include <stdio.h>

int main(void){
        printf("Random words");
}

使用命令 gcc -o print print.c 进行编译会导致完全干净的编译，并且可执行文件按预期运行。

使用命令 gcc -o -std=c99 print print.c 编译得到:

print: In function `_fini':
(.fini+0x0): multiple definition of `_fini'
/usr/lib/gcc/i686-linux-gnu/4.6.1/../../../i386-linux-gnu/crti.o:(.fini+0x0): first defined here
print: In function `__data_start':
(.data+0x0): multiple definition of `__data_start'
/usr/lib/gcc/i686-linux-gnu/4.6.1/../../../i386-linux-gnu/crt1.o:(.data+0x0): first defined here
print: In function `__data_start':
(.data+0x4): multiple definition of `__dso_handle'
/usr/lib/gcc/i686-linux-gnu/4.6.1/crtbegin.o:(.data+0x0): first defined here
print:(.rodata+0x4): multiple definition of `_IO_stdin_used'
/usr/lib/gcc/i686-linux-gnu/4.6.1/../../../i386-linux-gnu/crt1.o:(.rodata.cst4+0x0): first defined here
print: In function `_start':
(.text+0x0): multiple definition of `_start'
/usr/lib/gcc/i686-linux-gnu/4.6.1/../../../i386-linux-gnu/crt1.o:(.text+0x0): first defined here
print:(.rodata+0x0): multiple definition of `_fp_hw'
/usr/lib/gcc/i686-linux-gnu/4.6.1/../../../i386-linux-gnu/crt1.o:(.rodata+0x0): first defined here
print: In function `_init':
(.init+0x0): multiple definition of `_init'
/usr/lib/gcc/i686-linux-gnu/4.6.1/../../../i386-linux-gnu/crti.o:(.init+0x0): first defined here
/tmp/ccfvqWMW.o: In function `main':
print.c:(.text+0x0): multiple definition of `main'
print:(.text+0xb4): first defined here
/usr/lib/gcc/i686-linux-gnu/4.6.1/crtend.o:(.dtors+0x0): multiple definition of `__DTOR_END__'
print:(.dtors+0x4): first defined here
/usr/bin/ld: error in print(.eh_frame); no .eh_frame_hdr table will be created.
collect2: ld returned 1 exit status

有人知道如何解决这个问题吗？

Answer 1

您的 print 没有被解释为 -o 选项的参数，因此它试图从第一次编译。这包括许多重复它试图链接的其他符号的符号，因此会出现错误。

改为尝试:

gcc -std=c99 -o print print.c

或者更好:

gcc -std=c99 -Wall -g -O -o print print.c