‘main’通常是一个非静态函数ERROR

有两个错误出现：main.c：80：警告：’main’通常是一个非静态函数main.c：88：错误：输入结束时的预期声明或语句我似乎无法找到问题……花括号的数量相等……似乎有什么问题？

#include  #include  #include  #include  #include  #include "main-getopt.h" void print_usage_and_abort( const char *message ) { if( NULL != message ) fprintf( stderr, "Error: %s\n", message ); fprintf( stderr, "Usage: partitioner -n  [ -f  ]\n\n" ); exit( -1 ); } void parsing (int argc, char **argv, struct Params *params) { char error_message[256]; params->nodes = 0; memcpy( params->filename_base, "output", strlen("output") + 1 ); int opt; size_t len; int numarg; while ((opt = getopt(argc, argv, "n:f:")) != -1) { int i; for (i = 1; i < argc; i++) { if (argv[i][0] == '-') { if (i+1 == argc || argv[i+1][0] == '-') { sprintf( error_message, "No Filename"); print_usage_and_abort( error_message ); } if (argv[i][1] == 'n') { numarg = atoi( optarg ); if( numarg  2048 ) { sprintf( error_message, "Number of nodes agrument expects a " "number between 1 and 2048, actual %s", optarg ); print_usage_and_abort( error_message ); } } else if (argv[i][1] == 'f') len = strlen( optarg ); // limit to buffer capacity if( len >= MAX_FILENAME_BASE ) { sprintf( error_message, "Base filename parameter length is " "expected to be less than %d but is %d", (int)MAX_FILENAME_BASE, (int)len ); print_usage_and_abort( error_message ); } else if(lenfilename_base, optarg, len + 1 ); break; } else { sprintf( error_message, "Unknown command switch %c", (char)optopt ); print_usage_and_abort( error_message ); break; } } } if( 0==params->nodes ) { sprintf( error_message, "Number of nodes switch -n is required" ); print_usage_and_abort( error_message ); } } int main(int argc, char *argv[]) { struct Params params; parse_arguments( argc, argv, &params ); fprintf( stdout, "Parameters are:\n\tNumber of nodes:\t%d\n\t" "Filename base:\t%s\n\n", params.nodes, params.filename_base ); return 0; } 

Answer 1

我为你编辑了缩进。 你现在看到在parsingfunction的某个地方你错过了一个结束括号吗？